We are starting a five-day challenge focusing on key chemistry chapters for JEE Mains. Today's topic is chemical bonding, which is very scoring and has the potential for several exam questions.

Bond formation occurs when two atoms experience attractive and repulsive forces. As they approach each other, attractive forces dominate until a specific distance is reached, known as the bond formation point.

At this point, energy decreases and stability increases. If the atoms are brought closer than this point, repulsive forces increase, leading to higher energy and instability.

The potential energy curve for H2 molecules illustrates this concept. As the inter-nuclear distance decreases, energy decreases until bond formation occurs, defining the bond radius.

• ★The five-day challenge covers five important chemistry chapters for JEE Mains.0:25
• ★Bond formation occurs at a specific distance where attractive forces balance repulsive forces.1:00
• ★Energy decreases and stability increases until bond formation; further distance decrease leads to repulsion.2:00
• ★The potential energy curve for H2 molecules shows the relationship between inter-nuclear distance and energy.3:00
• ★The point of bond formation defines the minimum energy and maximum stability for the bond.4:00

Example: In J 2020, a direct question asked about the potential energy curve for H2 molecules based on inter-nuclear distance.4:00

There are two types of overlapping: head-on and sideways. Head-on overlapping results in a sigma bond, while sideways overlapping results in a pi or delta bond. In a sigma bond, two lobes overlap. In a pi bond, four lobes overlap, and in a delta bond, eight lobes overlap.

The strength of a sigma bond is greater than that of a pi bond. The order of overlapping strength is: $2s-2s<2p-2p<1s-1s$ for sigma bonds. For pi bonds, the order is $p\pi -p\pi <p\pi -d\pi <d\pi -d\pi$. Delta bonds have the most lobes overlapping, making them stronger than pi bonds.

• ★Head-on overlapping results in a sigma bond.4:42
• ★Delta bonds involve eight lobes overlapping.4:42
• ★Sigma bond strength is greater than pi bond strength.4:42
• ★Dpi-dpi overlapping is the strongest type of overlapping.4:42

Hybridization is the mixing of orbitals, such as S and P orbitals, to form new hybrid orbitals like sp, sp2, sp3, sp3d, sp3d2, and sp3d3. Exam questions frequently ask for the hybridization of given compounds. To determine the hybridization from a compound's formula, use the formula: $No.ofhybridorbitals=\frac{1}{2}\{\text{total no. of electrons in the central atom}+\text{total no. of monovalent atoms - charge}\}$.

Valence electrons for various groups are as follows: Group 13 has 3, Group 14 has 4, Group 15 has 5, Group 16 has 6, Group 17 has 7, and Group 18 has 8. This information is crucial for calculating hybridization.

• ★Hybridization involves mixing orbitals to form new hybrid orbitals.7:25
• ★Common hybridizations include sp, sp2, sp3, sp3d, sp3d2, and sp3d3.7:30
• ★Use the formula $No.ofhybridorbitals=\frac{1}{2}\{\text{total no. of electrons in the central atom}+\text{total no. of monovalent atoms - charge}\}$ to find hybridization.7:35
• ★Valence electrons for Group 13 to 18 are 3, 4, 5, 6, 7, and 8 respectively.8:00

To determine the hybridization of nitrogen, use the formula: total valence electrons + total number of monovalent atoms - charge on ion. For example, in $N{O}_2^{-}$, nitrogen has 5 valence electrons, there are no monovalent atoms, and the negative charge adds 1, leading to a calculation of $3$ for hybridization: $\frac{1}{2}[5+0+1]=3$.

In the case of $N_2^{+}$, nitrogen still has 5 valence electrons, but we subtract 1 for the positive charge, resulting in $4$ valence electrons. The hybridization is calculated as $\frac{1}{2}[5+0-1]=\frac{4}{2}=2$. For $N{H}_4^{+}$, nitrogen has 5 valence electrons, and there are 4 monovalent hydrogen atoms, leading to a calculation of $4$ monovalent atoms and a final hybridization of $3$: $\frac{1}{2}[5+4-1]$.

• ★Group number 15 has five valence electrons.9:53
• ★Monovalent atoms include hydrogen and elements from group 1 and group 17.9:53
• ★Positive charges subtract from the total count of valence electrons.9:53
• ★The hybridization formula is: total valence electrons + monovalent atoms - charge.9:53

Example: For $N{H}_4^{+}$, nitrogen has 5 valence electrons and 4 monovalent hydrogen atoms, leading to a hybridization of $3$.9:53

The steric number is calculated as the sum of sigma bonds and lone pairs. For example, if the steric number is 2, it corresponds to sp hybridization; if it is 3, it corresponds to sp2; and if it is 4, it corresponds to sp3. For steric numbers of 5 and 6, the hybridizations are sp3d and sp3d2, respectively.

In VSEPR theory, lone pair-lone pair interactions have higher repulsion compared to lone pair-bond pair interactions and bond pair-bond pair interactions. The VSEPR formula is given by the number of sigma bond pairs plus the number of lone pairs.

• ★Steric number is the sum of sigma bonds and lone pairs.12:26
• ★Steric number of 2 corresponds to sp hybridization.12:26
• ★Lone pair-lone pair interactions have higher repulsion.12:26
• ★VSEPR formula = No. of sigma bond pairs + No. of lone pairs.12:26

Example: For NH4+, the steric number is 4, leading to sp3 hybridization: $\frac{1}{2}[5+4-1]=\frac{8}{2}\Rightarrow s{p}^3$.12:26

In sp hybridization, the geometry is linear with a bond angle of $180°$. This means two atoms are connected in a straight line.

For sp2 hybridization, the geometry is trigonal planar with a bond angle of $120°$. This can be visualized as a triangle in a plane. In sp3 hybridization, the geometry is tetrahedral with bond angles of $109°$. The tetrahedral structure remains unchanged upon rotation.

• ★sp hybridization results in a linear geometry with a bond angle of $180°$.15:21
• ★sp2 hybridization has a trigonal planar geometry with a bond angle of $120°$.15:21
• ★Tetrahedral geometry is characterized by bond angles of $109°$.15:21
• ★sp3d hybridization is mentioned but not fully explained in this segment.15:21

Trigonal planar geometry is associated with sp2 hybridization. It forms a flat structure with three bonds. For sp3d hybridization, the geometry is trigonal bipyramidal, which includes two pyramids. This structure has five bonds, creating a bipyramidal shape.

Octahedral geometry consists of eight triangular faces. It can be visualized as a square planar arrangement with two additional bonds at 90° angles. The term 'octahedral' is preferred over 'square bipyramidal'. The steric number is calculated as the sum of sigma bonds and lone pairs, excluding pi bonds.

• ★Trigonal planar geometry is sp2 hybridized.17:21
• ★Octahedral geometry has eight triangular faces.17:21
• ★Steric number = number of sigma bonds + lone pairs.17:21
• ★Different hybridizations include sp, sp2, sp3, sp3d, and sp3d2.17:21

Example: Calculate the steric number for a molecule with 4 sigma bonds and 1 lone pair: $4+1=5$.17:21

Beryllium has four valence electrons, represented as $1s^{2}2s^2$. Since beryllium belongs to group number two, it has two valence electrons available for bonding. In bond formation, only the outermost valence electrons are involved; inner core electrons do not participate.

In the case of beryllium and chlorine, beryllium forms two bond pairs with chlorine, consuming one electron from each bond. Therefore, there are zero lone pairs in this scenario. Always calculate lone pairs, even if they may not be present, as this is crucial in exam situations.

• ★Beryllium has four valence electrons: $1s^{2}2s^2$20:04
• ★Beryllium belongs to group number two, indicating two valence electrons.20:08
• ★Only valence electrons participate in bond formation; core electrons do not.20:12
• ★Always calculate lone pairs, even if they are absent.20:20

Example: Beryllium forms two bond pairs with chlorine, consuming its two valence electrons.20:18

A lone pair of electrons is considered zero in hybridization calculations. For sp hybridization, two sigma bonds result in a linear geometry with a bond angle of 180°.

In BF3, boron has three valence electrons and forms three sigma bonds with fluorine, leading to sp2 hybridization and a trigonal planar geometry. CH4 exhibits sp3 hybridization due to four sigma bonds, resulting in a tetrahedral shape. NH4+ is analyzed similarly, confirming its sp3 hybridization and tetrahedral structure.

• ★A lone pair counts as zero in hybridization.22:11
• ★Two sigma bonds indicate sp hybridization with a 180° angle.22:15
• ★BF3 has sp2 hybridization and trigonal planar geometry.22:25
• ★CH4 and NH4+ both exhibit sp3 hybridization and tetrahedral shape.22:35

Example: In BF3, boron forms three sigma bonds with fluorine, leading to sp2 hybridization and a trigonal planar structure.22:25

To calculate the number of lone pairs, start with the valence electrons. For example, ammonia has a structure where one lone pair is donated to hydrogen, forming an ammoniated ion.

In PCl5, phosphorus has five valence electrons and forms five bonds with chlorine. The calculation is $5-5=0$, resulting in zero lone pairs. Thus, the hybridization is sp3d, leading to a trigonal bipyramidal structure.

For sulfur (S), which belongs to group 16, it has six valence electrons. In SF6, it forms six bonds, so $6-6=0$ lone pairs. The hybridization is sp3d2, corresponding to an octahedral shape.

• ★Lone pairs = valence electrons - bond pairs.24:10
• ★PCl5 has zero lone pairs and sp3d hybridization.24:15
• ★SF6 has zero lone pairs and sp3d2 hybridization.24:25
• ★Chlorine typically forms single bonds.24:30

Example: For PCl5, with five valence electrons and five bonds, the lone pairs are calculated as $5-5=0$.24:15

In this segment, we discuss the role of lone pairs in molecular geometry using sulfur dioxide (SO2) as an example. Sulfur, in group number 16, has six valence electrons. Oxygen prefers to form two bonds, leading to the formation of multiple bonds.

For SO2, after bonding, one lone pair is formed from the remaining electrons. The structure consists of two sigma bonds and one lone pair, resulting in an sp2 hybridization.

• ★Octahedral geometry is formed when lone pairs are zero.26:13
• ★Sulfur has six valence electrons and forms bonds with oxygen.26:13
• ★SO2 has two sigma bonds and one lone pair, leading to sp2 hybridization.26:13

Example: In SO2, one lone pair results from the remaining electrons after bonding.26:13

Molecular geometry considers both lone pairs and bond pairs, while molecular shape only considers bond pairs. When there are no lone pairs, the geometry and shape are the same. For example, in a trigonal planar geometry, if there is a lone pair, the shape becomes bent or V-shaped due to repulsion.

Ammonia (NH3) has one lone pair and three bond pairs. Nitrogen belongs to group 15, contributing five valence electrons. The steric number can be calculated to determine the molecular geometry.

• ★Molecular geometry includes lone pairs; shape does not.28:23
• ★Lone pairs cause repulsion, altering molecular shape.28:30
• ★Ammonia (NH3) has one lone pair and three bond pairs.28:45
• ★Calculate steric number for molecular geometry.28:55

Example: For NH3, nitrogen has five valence electrons and forms three bonds, resulting in one lone pair.28:50

To calculate molecular geometry, use the formula based on valence electrons. For a molecule with four valence electrons and one lone pair, the hybridization is $s{p}^3$, indicating tetrahedral geometry. However, when considering lone pairs, the shape becomes trigonal pyramidal due to the presence of one lone pair.

For example, in water (H2O), oxygen has six valence electrons. The calculation shows two lone pairs affecting the molecular shape. The formula used is $\frac{1}{2}[6+2]$, leading to $4=s{p}^3$. The lone pairs are calculated by subtracting occupied electrons from total valence electrons.

• ★Molecules with four valence electrons and one lone pair have trigonal pyramidal shape.31:04
• ★Neglect lone pairs when determining the molecular geometry.31:04
• ★Oxygen in H2O has six valence electrons contributing to its shape.31:04
• ★Lone pairs are calculated by subtracting occupied electrons from total valence electrons.31:04

Example: In H2O, the calculation shows two lone pairs affecting its geometry.31:04

To determine the number of lone pairs in a molecule, start with the total valence electrons. For example, in water, there are six valence electrons. Two are used in bonding with hydrogen, leaving four electrons, which means there are two lone pairs. This results in a bent molecular geometry.

For SF4, sulfur has six valence electrons, with four forming bonds with fluorine. This leaves one lone pair. The total number of electron pairs is five (4 sigma bonds + 1 lone pair). The hybridization of SF4 is sp3d. The lone pair is best positioned in the equatorial position to minimize repulsion, with equatorial angles at 120° and axial angles at 90°.

• ★Water has two lone pairs, resulting in a bent geometry.33:35
• ★SF4 has one lone pair and is sp3d hybridized.33:45
• ★Lone pairs in SF4 are best positioned in the equatorial position.33:55
• ★Equatorial angles in SF4 are 120°, axial angles are 90°.34:05

Example: In SF4, sulfur has six valence electrons, four bond pairs, and one lone pair, leading to sp3d hybridization.33:50

Lone pairs are always positioned in the equatorial position in trigonal bipyramidal geometry for stability. Ignoring lone pairs, the shape resembles a seesaw.

For CLF3, chlorine has 7 valence electrons. Three are used in bonding with fluorine, leaving 2 lone pairs. These lone pairs occupy equatorial positions, resulting in a T-shaped geometry. The hybridization for CLF3 is sp3 due to 3 bonding pairs and 2 lone pairs.

For BrF5, bromine also has 7 valence electrons. With 5 used in bonding, there is 1 lone pair. The presence of this lone pair leads to a hybridization of sp3d2.

• ★Lone pairs occupy equatorial positions in trigonal bipyramidal geometry for stability.35:35
• ★CLF3 has a T-shaped geometry due to 2 lone pairs in equatorial positions.35:45
• ★Hybridization for CLF3 is sp3 with 3 bonding pairs and 2 lone pairs.35:55
• ★BrF5 has 1 lone pair leading to sp3d2 hybridization.36:15

Example: For BrF5, 7 valence electrons minus 5 used in bonding leaves 1 lone pair, resulting in sp3d2 hybridization.36:25

A molecule with four bonds and one lone pair has a square pyramidal shape. For example, in XeF4, there are four bonding electrons and two lone pairs. Lone pairs are positioned opposite each other to minimize repulsion.

Ignoring lone pairs helps determine the molecular shape, resulting in a square planar structure. A central atom with two lone pairs and three single bonds has a total of five regions of electron density, leading to an sp3d hybridization.

• ★The shape of a molecule with four bonds and one lone pair is square pyramidal.38:13
• ★In XeF4, there are four bonding electrons and two lone pairs.38:25
• ★Lone pairs are positioned opposite each other to minimize repulsion.38:35
• ★A molecule with two lone pairs and three single bonds has sp3d hybridization.40:00

Example: A molecule with two lone pairs and three single bonds has a total of five regions of electron density, leading to sp3d hybridization.40:00

In a trigonal bipyramidal geometry, the presence of two lone pairs leads to a T-shaped molecular structure. The lone pairs are positioned to minimize repulsion, specifically in the equatorial positions.

For linear species, the steric number can be calculated to determine the molecular shape. For example, the N3 negative ion is linear, and its nitrogen hybridization can be calculated as $2.5$. Ozone (O3) has a sp2 hybridization, with one oxygen atom as the central atom and the other two as surrounding atoms.

• ★Trigonal bipyramidal with two lone pairs results in a T-shaped structure.40:47
• ★Steric number helps identify linear species.40:55
• ★N3 negative ion is linear with a hybridization of $2.5$.40:58
• ★Ozone (O3) has a sp2 hybridization.41:05

Example: The N3 negative ion is confirmed to be linear based on its structure.40:58

In this segment, we analyze the hybridization of nitrogen, which is sp2. Nitrogen has five valence electrons, and with an additional electron, it totals six. This leads to a bond order calculation.

The molecular shape discussed is linear, particularly for the N3 negative ion. Bond order is defined as the number of average bonds in a molecule, which can be fractional due to resonance. For example, oxygen has a bond order of two due to its double bond, while nitrogen can have a bond order of three due to its triple bond.

• ★Nitrogen has a hybridization of sp2.42:54
• ★Bond order can be fractional due to resonance.42:54
• ★Oxygen typically has a bond order of two.42:54
• ★Nitrogen can have a bond order of three.42:54

Bond order can be a fraction due to resonance, indicating the presence of partial double bonds. Bond order is directly proportional to bond strength; a higher bond order results in stronger bonds. Conversely, bond length is inversely proportional to bond order; thus, a higher bond order leads to shorter bond lengths.

Bond length is influenced by atomic size; larger atoms lead to longer bond lengths. As you move down a group in the periodic table, atomic size increases, resulting in increased bond lengths. Single bonds have the maximum bond length, followed by double bonds, and then triple bonds. Bond enthalpy is directly proportional to the number of bonds; more bonds require more energy to break. However, bond enthalpy is inversely proportional to atomic size; larger atoms form longer bonds that are easier to break.

• ★Bond order is defined as $1+\frac{\text{number of pi bonds}}{\text{number of sigma bonds}}$.45:05
• ★Bond length is directly related to atomic size; larger atoms lead to longer bonds.45:15
• ★Bond enthalpy is inversely proportional to atomic size; larger atoms have weaker bonds.45:25
• ★Fluorine has the smallest atomic size but does not have the highest bond enthalpy due to lone pair repulsions.45:35

Hybridization directly affects bond angles in molecules. For linear hybridization, the bond angle is $180^{\text{o}}$. In trigonal planar (sp2), the bond angle is $120^{\text{o}}$, and for tetrahedral (sp3), it is approximately $109.5^{\text{o}}$. In octahedral geometry, the bond angles are $90^{\text{o}}$.

When comparing molecules with the same hybridization, such as CH4, NH3, and H2O, the number of lone pairs influences the bond angles. CH4 has zero lone pairs, NH3 has one lone pair, and H2O has two lone pairs. The presence of lone pairs decreases bond angles due to repulsion between electron pairs, making the bond angle maximum in CH4.

• ★Hybridization affects bond angles in molecules.47:49
• ★Linear hybridization has a bond angle of $180^{\text{o}}$.47:55
• ★The presence of lone pairs decreases bond angles due to repulsion.48:15
• ★Bond angle is maximum in CH4 due to zero lone pairs.48:25

Example: In CH4, the bond angle is maximum at $109.5^{\text{o}}$ due to no lone pairs.48:20

The bond angles in molecules are influenced by lone pair repulsion. For example, in CH4, the bond angle is 109°, in NH3 it is 107°, and in H2O it is 104° due to the presence of lone pairs.

Lone pair-bond pair repulsion decreases bond angles as the number of lone pairs increases. The bond angle is directly proportional to the electronegativity of the central atom.

If hybridization and the number of lone pairs are the same, the bond angle is directly proportional to the size of surrounding atoms and inversely proportional to the electronegativity of surrounding atoms.

Among the compounds CH4, NH3, and H2O, CH3 has the highest bond angle due to having the least number of lone pairs. The order of bond angles can be determined by comparing electronegativities and hybridizations.

• ★Bond angles: CH4 = 109°, NH3 = 107°, H2O = 104° due to lone pair repulsion.49:54
• ★Lone pair-bond pair repulsion decreases bond angles as lone pairs increase.49:55
• ★Bond angle is directly proportional to the electronegativity of the central atom.49:56
• ★If hybridization and lone pairs are the same, bond angle is proportional to surrounding atom size.49:57
• ★CH3 has the highest bond angle due to the least number of lone pairs.49:58

Example: The order of bond angles can be determined as CH4 > H2O > NH3 based on electronegativity and lone pairs.49:59

The sp2 hybridization has a bond angle of 120 degrees. For PF3, phosphorus has five outermost electrons, three of which are occupied, resulting in one lone pair. Thus, PF3 exhibits sp3 hybridization. The hybridization order is BF3 > PF3 > ClF3 based on their structures.

Dipole moment is defined as the direction from less electronegative to more electronegative atoms. It is a vector quantity, and if the dipole moment (bc) is zero, the molecule is nonpolar. The direction of the dipole moment is always from positive to negative charge. If dipole moments cancel each other, the resultant dipole moment is zero. The angle between dipole moments affects the resultant value; larger angles lead to smaller resultant dipole moments.

• ★sp2 hybridization has a bond angle of 120 degrees.53:50
• ★PF3 has one lone pair and is sp3 hybridized.53:55
• ★If bc is zero, the molecule is nonpolar.54:10
• ★Dipole moment direction is from positive to negative charge.54:15

Example: In molecules where dipole moments cancel, the resultant dipole moment is zero.54:18

The resultant vector value decreases as the angle between two vectors increases. Therefore, if the angle, $\theta$, is larger, the resultant dipole moment, ${\text{mu}}_{\text{resultant}}$, is smaller. This means that a smaller angle results in a larger dipole moment.

To determine dipole moments, knowing the shape of the molecule is necessary. The shape is derived from hybridization. If you can draw the shape, you can answer questions about bond angles or dipole moments.

For example, among the molecules CH${}_4$, H${}_2$O, NH${}_3$, CCl${}_4$, and HCl, we identify those with nonzero dipole moments. Water (H${}_2$O) and ammonia (NH${}_3$) have significant dipole moments due to their structures, while BF${}_3$ and BeF${}_2$ have $\text{mu}=0$.

• ★Theta is inversely proportional to the angle between vectors.56:19
• ★Understanding molecular shape is key for determining dipole moments.56:25
• ★Water and ammonia have nonzero dipole moments due to their shapes.56:35
• ★$\mu =0$ for BF${}_3$ and BeF${}_2$; $\mu \ne 0$ for H${}_2$O, NH${}_3$, CCl${}_4$, and HCl.56:55

Example: Among CH${}_4$, H${}_2$O, NH${}_3$, CCl${}_4$, and HCl, three molecules have nonzero dipole moments.56:55

Molecular orbital theory states that valence electrons are associated with all nuclei in a molecule. Atomic orbitals combine to form molecular orbitals, which can be classified as bonding or anti-bonding. Bonding molecular orbitals have lower energy, while anti-bonding molecular orbitals are higher in energy. The stability of molecular orbitals is inversely proportional to their energy levels.

Bonding molecular orbitals are formed by the addition of wave functions, while anti-bonding molecular orbitals result from the subtraction of wave functions. Sigma bonding molecular orbitals can involve 2s and 1s orbitals, as well as 2pz orbitals. Pi bonding molecular orbitals are formed from 2px and 2py orbitals.

• ★Molecular orbital theory considers valence electrons associated with all nuclei.59:33
• ★Bonding molecular orbitals are lower in energy than anti-bonding molecular orbitals.59:45
• ★Bonding orbitals form by addition of wave functions; anti-bonding by subtraction.59:55
• ★Sigma bonding involves 2s, 1s, and 2pz orbitals; pi bonding involves 2px and 2py.60:05

Example: The angle of 120° can affect the resultant vector in molecular shapes.60:00

The 2p z orbital has sigma bonding, while the 2p x and 2p y orbitals have pi bonding, and they always have equal energies. For bonding molecular orbitals, we have sigma 1s, sigma 2s, sigma 2p x, pi 2p x, pi 2p y, and sigma 2p z. Anti-bonding orbitals are indicated with a star, such as sigma star 1s and pi star.

For electronic configurations with less than or equal to 14 electrons, the arrangement is $21212121$. For configurations with more than 14 electrons, the arrangement is $1221121$. In molecular orbital theory, sigma 1s and sigma star 1s are filled with a total of eight electrons.

Bond order is calculated using the formula: $\text{Bond Order}=\frac{1}{2}(\text{Number of electrons in bonding orbitals}-\text{Number of electrons in anti-bonding orbitals})$. For 14 electrons, the bond order is three; adding or removing an electron changes the bond order by $0.5$.

• ★2p z has sigma bonding; 2p x and 2p y have pi bonding with equal energies.61:51
• ★For configurations with ≤ 14 electrons, the arrangement is $21212121$.62:05
• ★Bond order is calculated as $\frac{1}{2}(\text{bonding electrons}-\text{anti-bonding electrons})$.62:15
• ★For 14 electrons, bond order is 3; changing electron count alters bond order by $0.5$.62:25

Paramagnetic substances have unpaired electrons, while diamagnetic substances have no unpaired electrons. An odd number of electrons indicates paramagnetism, and an even number typically indicates diamagnetism. However, there are exceptions, such as $O_2$ and $B_2$, which are paramagnetic despite having an even number of electrons.

To calculate bond order, we can use the number of electrons in a molecule. For $O_2^{-}$, $O_2$, and $O_2^{+}$, the electron counts are 17, 16, and 15 respectively. The bond order values are derived as follows: for 15 electrons, bond order is 2.5; for 16 electrons, it is 2; for 17 electrons, it is 1.5; and for 18 electrons, it is 1.

• ★Paramagnetic substances have unpaired electrons.64:46
• ★Diamagnetic substances have no unpaired electrons.64:55
• ★$O_2$ and $B_2$ are exceptions, being paramagnetic despite having even electrons.65:05
• ★Bond order can be calculated from the number of electrons in a molecule.65:15

Example: For 15 electrons, the bond order is 2.5; for 16, it is 2; for 17, it is 1.5; and for 18, it is 1.65:25

O2 has a maximum bond order and is a stable molecule. The bond order sequence is O2 > O2+ > O2- > O22-. O2 has 16 electrons. The filling of molecular orbitals follows the order: ${\text{σ}}_{1s}$, ${\text{σ}}_{1s}^{\ast }$, ${\text{σ}}_{2s}$, ${\text{σ}}_{2s}^{\ast }$, ${\text{σ}}_{2p_z}$, ${\text{π}}_{2p_x}$, ${\text{π}}_{2p_y}$, ${\text{π}}_{2p_x}^{\ast }$, ${\text{π}}_{2p_y}^{\ast }$, ${\text{σ}}_{2p_z}^{\ast }$. The electronic configuration for O2 is 1 2 2 1.

O2 has two unpaired electrons in the ${\text{π}}^{\ast }$ antibonding orbitals, specifically in ${\text{π}}_{2p_x}^{\ast }$ and ${\text{π}}_{2p_y}^{\ast }$. According to molecular orbital theory, a molecule with zero bond order will not exist. The bond order can be calculated using the formula: $\text{Bond Order}=\frac{\text{Number of bonding electrons}-\text{Number of antibonding electrons}}{2}$.

• ★O2 has a maximum bond order and is stable.66:48
• ★The electronic configuration for O2 is 1 2 2 1.66:55
• ★O2 has two unpaired electrons in the ${\text{π}}^{\ast }$ antibonding orbitals.67:05
• ★A molecule with zero bond order will not exist.67:15

Example: For helium, the bond order calculation shows it has 2 electrons in bonding orbitals.67:25

The viability of a molecule is determined by its electron configuration. For example, H2^2- has zero bonding electrons, making it unstable. Molecules with 0.5 bonding electrons are also very unstable.

Magnetism is classified into diamagnetic and paramagnetic. Diamagnetic substances have all electrons paired, while paramagnetic substances have at least one unpaired electron. The calculation of total electrons in various molecules, such as boron, is essential for determining their magnetic properties.

• ★Molecule H2^2- is unstable due to zero bonding electrons.69:11
• ★Diamagnetic substances have all electrons paired.69:11
• ★Paramagnetic substances have at least one unpaired electron.69:11
• ★Total electrons in a molecule can determine its magnetic properties.69:11

Example: For a certain molecule, the total number of electrons calculated is 14.69:11

Boron and oxygen are exceptions in paramagnetism. Generally, odd electron species are paramagnetic, while even electron species are diamagnetic. However, B2 and O2 are paramagnetic despite having an even number of electrons.

To determine which molecules exhibit paramagnetic behavior, we can use a simple trick based on the number of unpaired electrons. If a species contains unpaired electrons, it is paramagnetic. In the discussed examples, species A and D show paramagnetic behavior.

• ★Boron and oxygen are exceptions in paramagnetism.71:08
• ★Odd electron species are paramagnetic; even electron species are diamagnetic.71:08
• ★B2 and O2 are paramagnetic despite being even electron species.71:08
• ★Species A and D are paramagnetic based on unpaired electrons.71:08

Example: In the analysis of various molecules, species A and D are identified as paramagnetic due to their unpaired electrons.71:08

Molecules like E unprot, O2, and S2 are paramagnetic. This means they have unpaired electrons that contribute to their magnetic properties.

Sulfur has 16 electrons. This configuration leads to its paramagnetic nature, similar to oxygen. Cl2 has 34 electrons, which also results in paramagnetism.

B2, O2, and S2 are exceptions; they are paramagnetic despite having an even number of electrons. This is due to their electronic configurations.

Sulfur and oxygen exhibit similar properties because they belong to the same group in the periodic table, leading to comparable bonding characteristics.

• ★E unprot, O2, and S2 are paramagnetic due to unpaired electrons.73:36
• ★Sulfur has 16 electrons, contributing to its paramagnetic property.73:36
• ★Cl2 has 34 electrons and is also paramagnetic.73:36
• ★B2, O2, and S2 are paramagnetic despite having an even number of electrons.73:36
• ★Sulfur shows similar properties to oxygen because they are in the same group.73:36

Example: Cl2, with 34 electrons, is an example of a paramagnetic molecule.73:36