Chemical Bonding and Molecular Structure MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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81.
Among the following, the species having the smallest bond is
A
$$N{O^ - }$$
B
$$N{O^ + }$$
C
$${O_2}$$
D
$$NO$$
Answer :
$$N{O^ + }$$
$$\eqalign{
& N{O^ - }\left( {16} \right) - B.O. - 2 \cr
& {O_2}\left( {16} \right) - B.O. - 2 \cr
& N{O^ + }\left( {14} \right) - B.O. - 3 \cr
& NO\left( {15} \right) - B.O. - 2.5 \cr} $$
Higher the bond order lower is the bond length. Hence $$N{O^ + }$$ will have smallest bond.
82.
Which of the following is paramagnetic?
A
$$O_2^ - $$
B
$$C{N^ - }$$
C
$$CO$$
D
$$N{O^ + }$$
Answer :
$$O_2^ - $$
$$O_2^ - \left( {17{e^ - }} \right) - K\,K\,\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_x^2,\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\left\{ {{\pi ^*}2p_y^2 = {\pi ^*}2p_z^1} \right.} \right.$$
Thus, $$\,O_2^ - \,$$ has one unpaired electron; hence it is paramagnetic. Other species have no unpaired electron.
All of them have 14 electrons.
83.
Two elements $$X$$ and $$Y$$ combine to form a compound $$XY.$$ Under what conditions the bond formed between them will be ionic?
A
If the difference in electronegativities of $$X$$ and $$Y$$ is 1.9.
B
If the difference in electronegativities of $$X$$ and $$Y$$ is more than 1.9.
C
If the difference in electronegativities of $$X$$ and $$Y$$ is less than 1.9.
D
If both $$X$$ and $$Y$$ are highly electronegative.
Answer :
If the difference in electronegativities of $$X$$ and $$Y$$ is more than 1.9.
For the formation of an ionic bond, the electronegativities of two atoms should differ by more than 1.9.
84.
Predict the correct order among the following.
A
lone pair-lone pair > bond pair-bond pair > lone pair-bond pair
B
bond pair-bond pair > lone pair-bond pair > lone pair-lone pair
C
lone pair-bond pair > bond pair-bond pair > lone pair-lone pair
D
lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
According to the postulate of $$VSEPR$$ theory, a lone pair occupies more space than a bonding pair, since it lies closer to the central atom. This means that the repulsion between the different electron pairs follow the order.
$$lp - lp > lp - bp > bp - bp$$
85.
$$As{F_5}$$ molecule is trigonal bipyramidal. The hybrid orbitals used by $$As - {\text{atoms}}$$ for bonding are
A
$${d_{{x^2} - {y^2}}},{d_{{z^2}}},s,{p_x},{p_y}$$
B
$${d_{xy}},s,{p_x},{p_y},{p_z}$$
C
$$s,{p_x},{p_y},{p_z},{d_{xy}}$$
D
$${d_{{x^2} - {y^2}}},s,{p_x},{p_y},{p_z}$$
Answer :
$$s,{p_x},{p_y},{p_z},{d_{xy}}$$
\[As=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},\underbrace{3{{s}^{1}}3p_{x}^{1}\,\,3p_{y}^{1}\,\,3p_{z}^{1}\,\,3{{d}^{1}}}_{s{{p}^{3}}d\,\,\text{hybridisation}}\]
Due to this hybridisation, geometry of the $$As{F_5}$$ molecule is trigonal bipyramidal and the hybrid orbitals used by $$As - {\text{atoms}}$$ are $$s,{p_x},{p_y},{p_z}$$ and $${d_{xy}}.$$
86.
Oxygen molecule is paramagnetic because
A
no. of bonding electrons > no. of antibonding electrons
B
no. of bonding electrons < no. of antibonding electrons
C
no. of bonding electrons = no. of antibonding electrons
D
presence of unpaired electrons in molecular orbitals
Answer :
presence of unpaired electrons in molecular orbitals
87.
Order of size of $$sp,s{p^2}$$ and $$s{p^3}$$ orbitals is
A
$$s{p^3} < s{p^2} < sp$$
B
$$sp < s{p^2} < s{p^3}$$
C
$$s{p^2} < sp < s{p^3}$$
D
$$s{p^2} < s{p^3} < sp$$
Answer :
$$sp < s{p^2} < s{p^3}$$
The percentage of $$s$$ - character in $$s{p^3},s{p^2}$$ and $$sp$$ is 25%, 33% and 50% respectively. Order of size of orbitals is $$sp < s{p^2} < s{p^3}.$$
88.
In which of the following ionization processes the bond energy has increased and also the magnetic behaviour has changed from paramagnetic to diamagnetic ?
A
$$NO \to N{O^ + }$$
B
$${N_2} \to {N_2}^ + $$
C
$${C_2} \to {C_2}^ + $$
D
$${O_2} \to {O_2}^ + $$
Answer :
$$NO \to N{O^ + }$$
For $$NO$$
Total no. of electrons $$= 15$$
$$B.O = 2.5$$
$$Mag.$$ Behaviour = Paramagnetic
For $$N{O^ + }$$
Total no. of electrons $$= 14$$
$$B.O = 3$$
$$Mag.$$ Behaviour = Diamagnetic
89.
The internuclear distances in $$O – O$$ bonds for $$O_2^ + ,{O_2},O_2^ - $$ and $$O_2^{2 - }$$ respectively are :
A
$$1.30\,\mathop {\text{A}}\limits^{\text{o}} ,1.49\mathop {\text{A}}\limits^{\text{o}} ,1.12\mathop {\text{A}}\limits^{\text{o}} ,1.21\mathop {\text{A}}\limits^{\text{o}} $$
B
$$1.49\mathop {\text{A}}\limits^{\text{o}} ,1.21\mathop {\text{A}}\limits^{\text{o}} ,1.12\mathop {\text{A}}\limits^{\text{o}} ,1.30\mathop {\text{A}}\limits^{\text{o}} $$
C
$$1.21\mathop {\text{A}}\limits^{\text{o}} ,1.12\mathop {\text{A}}\limits^{\text{o}} ,1.49\mathop {\text{A}}\limits^{\text{o}} ,1.30\mathop {\text{A}}\limits^{\text{o}} $$
D
$$1.12\mathop {\text{A}}\limits^{\text{o}} ,1.21\mathop {\text{A}}\limits^{\text{o}} ,1.30\mathop {\text{A}}\limits^{\text{o}} ,1.49\mathop {\text{A}}\limits^{\text{o}} $$
The bond length follows the order
$$O_2^ + < {O_2} < O_2^ - < O_2^{2 - }$$
According to this the possible values are
$$1.12\mathop {\text{A}}\limits^{\text{o}} ,1.21\mathop {\text{A}}\limits^{\text{o}} ,1.30\mathop {\text{A}}\limits^{\text{o}} ,1.49\mathop {\text{A}}\limits^{\text{o}} $$
90.
The species in which the central atom uses $$s{p^2}$$ hybrid orbitals in its bonding is
A
$$P{H_3}$$
B
$$N{H_3}$$
C
$$CH_3^ + $$
D
$$Sb{H_3}$$
Answer :
$$CH_3^ + $$
From amongst given species $$P{H_3},$$ $$N{H_3}$$ and $$Sb{H_3}$$ are all $$s{p^3}$$ hybridised. Their central atom has both bond pair as well as lone pair of electrons. The lone pair occupy the fourth orbital. $$CH_3^ + $$ has only three pairs of electrons so it is $$s{p^2}$$ hybridised.