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The ion which is not tetrahedral in shape is
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1818 Upvotes
Verified Answer
The correct answer is:
$\mathrm{Cu}\left(\mathrm{NH}_3\right)_4{ }^{2+}$
$$
\begin{array}{ll}
\text { } & \text { }:\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+} \\
\mathrm{Cu}^{2+} \rightarrow 3 d^9 4 s^0
\end{array}
$$
$$
\text { One electron is shifted from } 3 d \text { to } 4 p \text { orbital. }
$$
\begin{array}{ll}
\text { } & \text { }:\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+} \\
\mathrm{Cu}^{2+} \rightarrow 3 d^9 4 s^0
\end{array}
$$
$$
\text { One electron is shifted from } 3 d \text { to } 4 p \text { orbital. }
$$
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