JEE MAIN - Chemistry (2023 - 8th April Morning Shift - No. 8)
In chromyl chloride, the number of d-electrons present on chromium is same as in
(Given at no. of $$\mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26$$ )
(Given at no. of $$\mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26$$ )
Mn (VII)
Ti (III)
Fe (III)
V (IV)
Explanation
When a mixture containing chloride ion is heated with $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ and concentrated $\mathrm{H}_2 \mathrm{SO}_{4^{\prime}}$ deep orange-red fumes of chromyl chloride $\left(\mathrm{CrO}_2 \mathrm{Cl}_2\right)$ are formed
$$ \begin{aligned} & \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+4 \mathrm{NaCl}+6 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{KHSO}_4+4 \mathrm{NaHSO}_4 +2 \mathrm{CrO}_2 \mathrm{Cl}_2 \uparrow+3 \mathrm{H}_2 \mathrm{O} \end{aligned} $$
Orange-red fumes
So in this case, $\mathrm{X}$ in $\mathrm{CrO}_2 \mathrm{Cl}_2$.
Oxidation state of $\mathrm{Cl}=-1, \mathrm{O}=-2, \mathrm{Cr}=x$
$$ x+2 \times(-2)+2 \times(-1)=0 \Rightarrow x=+6 $$
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$$ \begin{aligned} & \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+4 \mathrm{NaCl}+6 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{KHSO}_4+4 \mathrm{NaHSO}_4 +2 \mathrm{CrO}_2 \mathrm{Cl}_2 \uparrow+3 \mathrm{H}_2 \mathrm{O} \end{aligned} $$
Orange-red fumes
So in this case, $\mathrm{X}$ in $\mathrm{CrO}_2 \mathrm{Cl}_2$.
Oxidation state of $\mathrm{Cl}=-1, \mathrm{O}=-2, \mathrm{Cr}=x$
$$ x+2 \times(-2)+2 \times(-1)=0 \Rightarrow x=+6 $$
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