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In which of the following pairs both the ions are coloured in aqueous solutions ?
Options:
Solution:
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Verified Answer
The correct answer is:
$\mathrm{Ni}^{2+}, \mathrm{Ti}^{3+}$
$\mathrm{Sc}^{3+}(\mathrm{Z}=18): 1 s^2, 2 s^2 p^6, 3 s^2 p^6 d^0, 4 s^0$; no unpaired electron.
$\mathrm{Cu}^{+}(\mathrm{Z}=28): 1 s^2, 2 s^2 p^6, 3 s^2 p^6 d^{10}, 4 s^0$; no unpaired electron.
$$
\mathrm{Ni}^{2+}(\mathrm{Z}=26): 1 s^2, 2 s^2 p^6, 3 s^2 p^6 d^8, 4 s^0
$$
unpaired electrons are present.
$$
\mathrm{Ti}^{3+}(\mathrm{Z}=19): 1 s^2, 2 s^2 p^6, 3 s^2 p^6 d^1, 4 s^0 \text {; }
$$
unpaired electron is present
$$
\mathrm{Co}^{2+}(\mathrm{Z}=25): 1 s^2, 2 s^2 p^6, 3 s^2 p^6 d^7, 4 s^0 \text {; }
$$
unpaired electrons are present
So from the given options the only correct combination is $\mathrm{Ni}^{2+}$ and $\mathrm{Ti}^{3+}$.
$\mathrm{Cu}^{+}(\mathrm{Z}=28): 1 s^2, 2 s^2 p^6, 3 s^2 p^6 d^{10}, 4 s^0$; no unpaired electron.
$$
\mathrm{Ni}^{2+}(\mathrm{Z}=26): 1 s^2, 2 s^2 p^6, 3 s^2 p^6 d^8, 4 s^0
$$
unpaired electrons are present.
$$
\mathrm{Ti}^{3+}(\mathrm{Z}=19): 1 s^2, 2 s^2 p^6, 3 s^2 p^6 d^1, 4 s^0 \text {; }
$$
unpaired electron is present
$$
\mathrm{Co}^{2+}(\mathrm{Z}=25): 1 s^2, 2 s^2 p^6, 3 s^2 p^6 d^7, 4 s^0 \text {; }
$$
unpaired electrons are present
So from the given options the only correct combination is $\mathrm{Ni}^{2+}$ and $\mathrm{Ti}^{3+}$.
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