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
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 ?
$$NaF$$ 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
$$\sigma _b^2\sigma _a^{_ * 2}\sigma _b^2\sigma _a^{^ * 2}\left( {\pi _b^2 = \pi _b^2} \right)\sigma _b^1\left( {N_2^ + = 13\,{\text{electrons}}} \right)$$
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 $$ = 1.60 \times {10^{ - 19}}C$$ )
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
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.