Q: Draw molecular orbital diagram of \${O}_{2}\$ or \${N}_{2}\$ with magnetic behavior and bond order.

As it can be seen from the MOT of \$O_2\$, The electrons in the highest occupied molecular orbital are unpaired therefore it is paramagnetic in nature. Also, the bond order can be calculated as \$[N_b-N_a] / 2 = [10-6] / 2 = 2\$. Therefore there is a double bond present as \$O=O\$.

Q: In \${ PO }_{ 4 }^{ 3- }\$, the formal charge on each oxygen atom and the \$P-O\$ bond order respectively are:

In \${ PO }_{ 4 }^{ 3- }\$, the formal charge on each oxygen atom and the \$P-O\$ bond order are \$-0.75,1.25\$ respectively.\$\left[O-\underset{O}{\underset{||}{\overset{O}{\overset{|}{P}}}}-O\right]^{3-} \leftrightarrow \left[O-\underset{P}{\underset{|}{\overset{O}{\overset{|}{P}}}}=O\right]^{3-}\$\$\left[O-\underset{O}{\underset{|}{\overset{O}{\overset{||}{P}}}}-O\right]^{3-} \leftrightarrow \left[O=\underset{P}{\underset{|}{\overset{O}{\overset{|}{P}}}}-O\right]^{3-}\$Bond order \$=\cfrac { \text { Number of bonds} }{ \text {Number of Resonating structures} } =\cfrac { 5 }{ 4 } =1.25\$In a given resonance structure, the O atom that forms double bond has formal charge of 0 and the remaining 3 O atoms have formal charge of -1 each.In the resonance hybrid, a total of -3 charge is distributed over 4 O atoms. Thus the formal charge of each O atom is \$\dfrac {-3}{4} = -0.75\$.Note: To calculate the formal charge, the following formula is used.Formal Charge \$=\$ [Number of valence electrons on atom] \$–\$ [non-bonded electrons \$+\$ number of bonds]For O atom that forms double bond with P atom,Formal Charge \$=\$ 6] \$–\$ [4 \$+\$2] \$=\$ 0.For O atom that forms single bond with P atom,Formal Charge \$=\$ 6] \$–\$ [6 \$+\$1] \$=-\$ 1.

Q: Bond order of \$O_2, O_2^+, O_2^-\$ and \$O_2^{2-}\$ is in order________.

Bond order of the following is: \${ O }_{ 2 }^{ 2- }=1, { O }_{ 2 }^{ - }=1.5,{ O }_{ 2 }=2, { O }_{ 2 }^{ + }=2.5\$

Q: Which out of \$ \displaystyle NH_{3} \$ and \$ \displaystyle NF_{3} \$ has higher dipole moment and why?

The dipole moment of ammonia \$(1.47D)\$ is higher than the dipole moment of \$\displaystyle NF_3\$ \$(0.24D)\$. The molecular geometry is pyramidal for both the molecules. In each molecule, N atom has one lone pair. F is more electronegative than H and \$N-F\$ bond is more polar than \$N-H\$ bond. Hence, \$\displaystyle NF_3\$ is expected to have much larger dipole moment than \$\displaystyle NH_3\$. However reverse is true as in case of ammonia, the direction of the lone pair dipole moment and the bond pair dipole moment is same whereas in case of \$\displaystyle NF_3\$ it is opposite. Thus, in ammonia molecule, individual dipole moment vectors add whereas in \$\displaystyle NF_3\$, they cancel each other.

Q: Which of the following would have a permanent dipole moment?

Permanent dipole moment :A: \$B{ F }_{ 3 }\$(Image 1)In \$B{ F }_{ 3 }\$, the dipole moments cancel each other. Hence it does not have a permanent dipole moment.B.\$S{ F }_{ 4 }\$(Image 2)The hybridization of a \$S{ F }_{ 4 }\$ molecule is \$s{ p }^{ 3 }d\$, thus a seesaw structure (bent because of lp-lp repulsion) and thus having a net dipole moment.C.\$Si{ F }_{ 4 }\$(Image 3)In \$Si{ F }_{ 4 }\$ the dipole moments cancel each other. Hence it does not have a permanent dipole moment.D. \$Xe{ F }_{ 4 }\$-has a planar structure, so it is non-polar(Image 4)Hence the correct option is B - \$S{ F }_{ 4 }\$ - has a permanent dipole.

Q: Compare the relative stability of the following species and indicate their magnetic properties :\$ \displaystyle O_{2},\ O^{+}_{2},\ {O_2}^- \$ (superoxide) and \$ \displaystyle {O_2}^{2-} \$ (peroxide).

(1) The electronic configuration of \$\displaystyle O_2\$ molecule is \$\displaystyle KK(\sigma 2s )^2(\sigma ^*2s )^2(\sigma 2p_z )^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi ^* 2p_x )^1(\pi ^* 2p_y )^1 \$Its bond order is \$\displaystyle \frac {8-4}{2} = 2\$. It contains 2 unpaired electrons and is paramagentic.(2) The electronic configuration of \$\displaystyle O_2^+\$ ion is \$\displaystyle KK(\sigma 2s )^2(\sigma ^*2s )^2(\sigma 2p_z )^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi ^* 2p_x )^1\$Its bond order is \$\displaystyle \frac {8-3}{2} = 2.5\$. It contains 1 unpaired electron and is paramagnetic.(3) The electronic configuration of \$\displaystyle O_2^-\$ ion is \$\displaystyle KK(\sigma 2s )^2(\sigma ^*2s )^2(\sigma 2p_z )^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi ^* 2p_x )^2(\pi ^* 2p_y )^1 \$Its bond order is \$\displaystyle \frac {8-5}{2} = 1.5\$. It contains 1 unpaired electron and is paramagnetic.(4) The electronic configuration of \$\displaystyle O_2^{2-}\$ ion is \$\displaystyle KK(\sigma 2s )^2(\sigma ^*2s )^2(\sigma 2p_z )^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi ^* 2p_x )^2(\pi ^* 2p_y )^2 \$Its bond order is \$\displaystyle \frac {8-6}{2} = 1\$. It contains all paired electrons and is diamagnetic.Higher is the bond order, stronger is the bond, and higher is the stability.The decreasing order of the stability is \$\displaystyle O_2^+ > O_2 >{O_2}^- > O_2^{2-} \$.

Question 1

Draw molecular orbital diagram of ${O}_{2}$ or ${N}_{2}$ with magnetic behavior and bond order.

Accepted Answer

As it can be seen from the MOT of $O_2$, The electrons in the highest occupied molecular orbital are unpaired therefore it is paramagnetic in nature. Also, the bond order can be calculated as $[N_b-N_a] / 2 = [10-6] / 2 = 2$. Therefore there is a double bond present as $O=O$.

Question 2

In ${ PO }_{ 4 }^{ 3- }$, the formal charge on each oxygen atom and the $P-O$ bond order respectively are:

Accepted Answer

In ${ PO }_{ 4 }^{ 3- }$, the formal charge on each oxygen atom and the $P-O$ bond order are $-0.75,1.25$ respectively.$\left[O-\underset{O}{\underset{||}{\overset{O}{\overset{|}{P}}}}-O\right]^{3-} \leftrightarrow&#10; &#10;\left[O-\underset{P}{\underset{|}{\overset{O}{\overset{|}{P}}}}=O\right]^{3-}$$\left[O-\underset{O}{\underset{|}{\overset{O}{\overset{||}{P}}}}-O\right]^{3-} \leftrightarrow&#10; &#10;\left[O=\underset{P}{\underset{|}{\overset{O}{\overset{|}{P}}}}-O\right]^{3-}$Bond order $=\cfrac { \text { Number  of bonds} }{ \text {Number of  Resonating structures} } =\cfrac { 5 }{ 4 } =1.25$In a given resonance structure, the O atom that forms double bond has formal charge of 0 and the remaining 3 O atoms have formal charge of -1 each.In the resonance hybrid, a total of -3 charge is distributed over 4 O atoms. Thus the formal charge of each O atom is $\dfrac {-3}{4} = -0.75$.Note: To calculate the formal charge, the following formula is used.Formal Charge $=$ [Number of valence electrons on atom] $&#8211;$ [non-bonded electrons $+$ number of bonds]For O atom that forms double bond with P atom,Formal Charge $=$ 6] $&#8211;$ [4 $+$2] $=$ 0.For O atom that forms single bond with P atom,Formal Charge $=$ 6] $&#8211;$ [6 $+$1] $=-$ 1.

Question 3

Bond order of $O_2, O_2^+, O_2^-$ and $O_2^{2-}$ is in order________.

Accepted Answer

Bond order of the following is: ${ O }_{ 2 }^{ 2- }=1, { O }_{ 2 }^{ - }=1.5,{  O }_{ 2 }=2, { O }_{ 2 }^{ + }=2.5$

Question 4

What are hybridisation states of each carbon atom in the following compounds?$CH_2=C=O,\, CH_3CH=CH_2,\, (CH_3)_2CO,\, CH_2=CHCN,\, C_6H_6$

Accepted Answer

The hybridisation states of each carbon atom in the following compounds are given below. $\underset {sp^2}{C}H_2=\underset {sp}{C}=O,\, \underset {sp^3}{C}H_3\underset {sp^2}{C}H=\underset {sp^2}{C}H_2,\, (\underset {sp^3}{C}H_3)_2\underset {sp^2}{C}O,\, \underset {sp^2}{C}H_2=\underset {sp^2}{C}H\underset {sp^3}{C}N,\, \underset {sp^2}{C_6}H_6$

Question 5

Which out of $ \displaystyle NH_{3} $ and $ \displaystyle NF_{3} $ has higher dipole moment and why?

Accepted Answer

The dipole moment of ammonia $(1.47D)$ is higher than the dipole moment of $\displaystyle NF_3$ $(0.24D)$. The molecular geometry is pyramidal for both the molecules. In each molecule, N atom has one lone pair. F is more electronegative than H and $N-F$ bond is more polar than $N-H$ bond. Hence, $\displaystyle NF_3$ is expected to have much larger dipole moment than $\displaystyle NH_3$. However reverse is true as in case of ammonia, the direction of the lone pair dipole moment and the bond pair dipole moment is same whereas in case of $\displaystyle NF_3$ it is opposite. Thus, in ammonia molecule, individual dipole moment vectors add whereas in $\displaystyle NF_3$, they cancel each other.

Question 6

Define hydrogen bond. Explain different types of hydrogen bond giving examples.

Accepted Answer

Hydrogen bond is a electrostatic attraction between a hydrogen atom which is bond to a more electronegative atom such as Nitrogen, Oxygen, fluorine.These are two types of hydrogen bonds :-1) Intermolecular Hydrogen bonding :- It occurs between two separate molecules.Eq :- $H-F......H-F......H-F$                         $\updownarrow $                       $H-$bonding (Intermolecular)2) Intermolecular Hydrogen bonding :- It occurs among the different facts of same molecule.Eq :-

Question 7

Which of the following would have a permanent dipole moment?

Accepted Answer

Permanent dipole moment :A: $B{ F }_{ 3 }$(Image 1)In $B{ F }_{ 3 }$, the dipole moments cancel each other. Hence it does not have a permanent dipole moment.B.$S{ F }_{ 4 }$(Image 2)The hybridization of a $S{ F }_{ 4 }$ molecule is $s{ p }^{ 3 }d$, thus a seesaw structure (bent because of lp-lp repulsion) and thus having a net dipole moment.C.$Si{ F }_{ 4 }$(Image 3)In $Si{ F }_{ 4 }$ the dipole moments cancel each other. Hence it does not have a permanent dipole moment.D. $Xe{ F }_{ 4 }$-has a planar structure, so it is non-polar(Image 4)Hence the correct option is B - $S{ F }_{ 4 }$ - has a permanent dipole.

Question 8

Compare the relative stability of the following species and indicate their magnetic properties :$ \displaystyle O_{2},\ O^{+}_{2},\ {O_2}^-  $ (superoxide) and $ \displaystyle {O_2}^{2-}  $ (peroxide).

Accepted Answer

(1) The electronic configuration of $\displaystyle O_2$ molecule is $\displaystyle KK(\sigma 2s )^2(\sigma ^*2s )^2(\sigma 2p_z )^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi ^* 2p_x )^1(\pi ^* 2p_y )^1 $Its bond order is $\displaystyle \frac {8-4}{2} = 2$. It contains 2 unpaired electrons and is paramagentic.(2) The electronic configuration of $\displaystyle O_2^+$ ion is $\displaystyle KK(\sigma 2s )^2(\sigma ^*2s )^2(\sigma 2p_z )^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi ^* 2p_x )^1$Its bond order is $\displaystyle \frac {8-3}{2} = 2.5$. It contains 1 unpaired electron and is paramagnetic.(3) The electronic configuration of $\displaystyle O_2^-$ ion is $\displaystyle KK(\sigma 2s )^2(\sigma ^*2s )^2(\sigma 2p_z )^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi ^* 2p_x )^2(\pi ^* 2p_y )^1 $Its bond order is $\displaystyle \frac {8-5}{2} = 1.5$. It contains 1 unpaired electron and is paramagnetic.(4) The electronic configuration of $\displaystyle O_2^{2-}$ ion is $\displaystyle KK(\sigma 2s )^2(\sigma ^*2s )^2(\sigma 2p_z )^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi ^* 2p_x )^2(\pi ^* 2p_y )^2 $Its bond order is $\displaystyle \frac {8-6}{2} = 1$. It contains all paired electrons and is diamagnetic.Higher is the bond order, stronger is the bond, and higher is the stability.The decreasing order of the stability is $\displaystyle O_2^+ > O_2 >{O_2}^- > O_2^{2-} $.

Question 9

Determine the shape of the following molecules using the VSEPR model.$ \displaystyle BeCl_{2},\ BCl_{3},\ SiCl_{4},\ AsF_{5},\ H_{2}S \ and \ PH_{3} $

Accepted Answer

The shape according to VSEPR theory is given below : MoleculeNumber of electron pairs around central atom Molecular geometry Bond angles  $\displaystyle BeCl_2 $ 2 Linear $\displaystyle 180^o$ $\displaystyle  BCl_3$ 3 trigonal planar $\displaystyle 120^o$  $\displaystyle  SiCl_4$ 4 tetrahedral  $\displaystyle 109.5^o$ $\displaystyle  AsF_5$ 5 trigonal bipyramidal three $\displaystyle 120^o$, two  $\displaystyle 90^o$ $\displaystyle  H_2S$ 6non linear/bent  $\displaystyle 92^o$ $\displaystyle  PH_3$ 5 trigonal pyramidal  $\displaystyle 93.5^o$

Question 10

What is meant by the term bond order? Calculate the bond order of: $N_2, O_2, O^+_2$ and $O^-_2$?

Accepted Answer

Bond order: The number of chemical bonds present between two atoms of a molecule.$Bond\quad order=\dfrac { \left[ No.\quad of\quad electrons\quad in\quad bonding\quad orbital \right] -\left[ No.\quad of\quad electrons\quad in\quad anti-bonding\quad orbital \right]  }{ 2 } $In ${ N }_{ 2 }\Rightarrow Bond\quad order=\dfrac { 10-4 }{ 2 } =3\Rightarrow N\equiv N$ In $O_{ 2 }\Rightarrow Bond\quad order=\dfrac { 10-6 }{ 2 } =2\Rightarrow O\equiv O$In ${ O }_{ 2 }^{ + }$, no. of electrons in bonding orbital is $9$$Bond\quad order=\dfrac { 9-6 }{ 2 } =\dfrac { 3 }{ 2 } =1.5$In ${ O }_{ 2 }^{ - }$, no. of electrons in anti-bonding orbital is $7$$Bond\quad order=\dfrac { 10-7 }{ 2 } =\dfrac { 3 }{ 2 } =1.5$