Chemical Bonding and Molecular Structure MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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221.
Which of the following is a polar molecule?
A
$$B{F_3}$$
B
$$S{F_4}$$
C
$$Si{F_4}$$
D
$$Xe{F_4}$$
Answer :
$$S{F_4}$$
Symmetrical molecules are generally non-polar although they have polar bonds. This is because bond dipole of one bond is cancelled by that of the other. $$B{F_3},Si{F_4}\,{\text{and}}\,Xe{F_4}$$ being symmetrical as non-polar. $$S{F_4}$$ is unsymmetrical because of the presence of a lone pair of electrons. Due to which it is a polar molecule.
222.
In $$Xe{F_2},Xe{F_4}$$ and $$Xe{F_6},$$ the number of lone pairs on $$Xe$$ are respectively
A
2, 3, 1
B
1, 2, 3
C
4, 1, 2
D
3, 2, 1
Answer :
3, 2, 1
In $$Xe{F_2}$$ Total number of valence electrons of $$Xe = 8,$$ two electrons shared with $$2F$$ atoms, 6 electrons left hence 3 lone pairs, in $$Xe{F_4}4$$ shared with $$4 F$$ atoms 4 left hence 2 lone pairs; in $$Xe{F_6}6$$ shared with $$6 F$$ atoms 2 left hence 1 lone pair.
223.
Among the following which compound will show the highest lattice energy ?
A
$$KF$$
B
$$NaF$$
C
$$CsF$$
D
$$RbF$$
Answer :
$$NaF$$
$$NaF$$ has high lattice energy because $$N{a^ + }$$ is smallest in size and lattice energy increases as the size of cation decreases. ( In the given question anion is common in all compound )
224.
Which of these statements is not true?
A
$$N{O^ + }$$ is not isoelectronic with $${O_2}$$
B
$$B$$ is always covalent in its compounds
C
In aqueous solution, the $$T{l^ + }\,ion$$ is much more stable than $$Tl\left( {{\rm{III}}} \right)$$
D
$$LiAl{H_4}$$ is a versatile reducing agent in organic synthesis.
Answer :
$$N{O^ + }$$ is not isoelectronic with $${O_2}$$
$$\left( {\text{A}} \right)N{O^ + } = 7 + 8 - 1 = 14\,{e^ - }.$$
$${O_2} = 16\,{e^ - }$$
i.e not iso-electronic
(B) Boron forms only covalent compounds. This is due to its extremely high ionisation energy.
(C) Compounds of $$T{l^ + }$$ are much more stable than those of $$T{l^{3 + }}.$$
(D) $$LiAl{H_4}$$ is a versatile reducing agent in organic synthesis
225.
$${N_2}$$ and $${O_2}$$ are converted into monocations, $$N_2^ + $$ and $$O_2^ + $$ respectively. Which of the following statements is wrong?
A
In $$N_2^ + ,N - N$$ bond weakens
B
In $$O_2^ + ,$$ the $$O - O$$ bond order increases
C
In $$O_2^ + ,$$ paramagnetism decreases
D
$$N_2^ + $$ becomes diamagnetic
Answer :
$$N_2^ + $$ becomes diamagnetic
$$\sigma _b^2\sigma _a^{_ * 2}\sigma _b^2\sigma _a^{^ * 2}\left( {\pi _b^2 = \pi _b^2} \right)\sigma _b^1\left( {N_2^ + = 13\,{\text{electrons}}} \right)$$
it contains one unpaired electron hence paramagnetic
226.
Bond distance in $$HF$$ is $$9.17 \times {10^{ - 11}}m.$$ Dipole moment of $$HF$$ is $$6.104 \times {10^{ - 30}}Cm.$$ The percentage ionic character in $$HF$$ will be :
( electron charge $$ = 1.60 \times {10^{ - 19}}C$$ )
227.
The conditions for the combination of atomic orbitals to form molecular orbitals are stated below. Mark the incorrect condition mentioned here.
A
The combining atomic orbitals must have nearly same energy.
B
The combining atomic orbitals must overlap to maximum extent.
C
Combining atomic orbitals must have same symmetry about the molecular axis.
D
$$Pi\left( \pi \right)$$ molecular orbitals are symmetrical around the bond axis.
Answer :
$$Pi\left( \pi \right)$$ molecular orbitals are symmetrical around the bond axis.
$$Sigma$$ molecular orbitals are symmetrical around the bond axis while $$pi$$ molecular orbitals are not symmetrical.
228.
Which of the following pairs will form the most stable ionic bond?
A
$$Na\,\,{\text{and}}\,Cl$$
B
$$Mg\,\,{\text{and}}\,F$$
C
$$Li\,\,{\text{and}}\,F$$
D
$$Na\,\,{\text{and}}\,F$$
Answer :
$$Mg\,\,{\text{and}}\,F$$
The ionic bond between $$Mg$$ and $$F$$ is most stable because in these the electrostatic force of attraction is maximum. As $$Mg$$ has high electropositive character and $$F$$ has high electronegative character among all other options that are given in question.
229.
The decreasing values of bond angles from $$N{H_3}\left( {{{106}^ \circ }} \right)\,{\text{to}}\,Sb{H_3}\left( {{{101}^ \circ }} \right)$$ down group-15 of the periodic table is due to
A
decreasing $$lp{\text{ - }}bp$$ repulsion
B
decreasing electronegativity
C
increasing $$bp{\text{ - }}bp$$ repulsion
D
increasing p - orbital character in $$s{p^3}$$
Answer :
decreasing electronegativity
The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.
\[\begin{matrix}
N{{H}_{3}} \\
{{107}^{\circ }} \\
\end{matrix}\,\,\begin{matrix}
P{{H}_{3}} \\
{{94}^{\circ }} \\
\end{matrix}\,\,\begin{matrix}
AS{{H}_{3}} \\
{{92}^{\circ }} \\
\end{matrix}\,\,\begin{matrix}
Sb{{H}_{3}} \\
{{91}^{\circ }} \\
\end{matrix}\,\,\begin{matrix}
Bi{{H}_{3}} \\
{{90}^{\circ }} \\
\end{matrix}\]
NOTE : This can also be explained by the fact that as
the size of central atom increases $$s{p^3}$$ hybrid orbital becomes more distinct with increasing size of central atom i.e, pure $$p$$ - orbitals are utilized in $$M - H$$ bonding
230.
Given below is the table showing shapes of some molecules having lone pairs of electrons. Fill up the blanks left in it.
Molecule type
$$bp$$
$$lp$$
Shape
Example
$$A{B_2}{E_2}$$
2
$$\underline {\,\,P\,\,} $$
Bent
$${H_2}O$$
$$A{B_3}{E_2}$$
3
2
$$\underline {\,\,Q\,\,} $$
$$Cl{F_3}$$
$$A{B_5}E$$
5
$$\underline {\,\,R\,\,} $$
$$\underline {\,\,S\,\,} $$
$$Br{F_5}$$
$$A{B_4}{E_2}$$
4
2
$$\underline {\,\,T\,\,} $$
$$\underline {\,\,U\,\,} $$
$$P$$
$$Q$$
$$R$$
$$S$$
$$T$$
$$U$$
(a)
2
Square pyramidal
2
T-shaped
Squar planar
$${H_2}{O_2}$$
(b)
4
T-shaped
5
Square planar
Square pyramidal
$$S{O_3}$$
(c)
2
T-shaped
1
Square pyramidal
Square planar
$$Xe{F_4}$$
(d)
3
Square planar
2
T-shaped
Square pyramidal
$$BrC{l_3}$$
A
(a)
B
(b)
C
(c)
D
(d)
Answer :
(c)
No explanation is given for this question. Let's discuss the answer together.