Explanation

(c)
$\left\lbrack Cr\left( NH_{3} \right)_{6} \right\rbrack\ \ \ \ \ \ \ \ \ \ Cl_{3} \rightleftharpoons$
Coordinate sphere Ionisable
$\left\lbrack Cr\left( NH_{3} \right)_{6} \right\rbrack^{3 +} + 3Cl^{-}\overset{AgNO_{3}}{\rightarrow}AgCl \downarrow$
white precipitate
$Cr(24)\lbrack Ar\rbrack 3d^{5}\ \ \ 4s^{1}$
${Cr}^{3 +}\lbrack Ar\rbrack 3d^{3}\ \ \ \ \ \ \ 4s^{0}$
3d 4s 4p
[table] $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | | | | | | | | [/table]
$Cr^{3 +}$ [Ar]
$\left\lbrack Cr\left( NH_{3} \right)_{6} \right\rbrack^{3 +}$ $\lbrack Ar\rbrack$ [image:9_image277.emf]
$\boxed{\times}$ Indicates lone-pair of $NH_{3}$ donated to Cr
$d^{2}sp^{3}$ hybridisation, octahedral, thus, correct.
There are three unpaired electrons, hence paramagnetic, thus correct.
$d^{2}sp^{3}$-inner orbital complex, thus incorrect
$Due\ to\ ionisable\ Cl^{-}$ions, white precipitate with $AgNO_{3}$, thus correct.
Therefore, (c) is wrong.