- AAlong the $x-$ axis
- BAlong the $y-$ axis
- ✓At an angle of ${45^o}$ from the $x$ and $y-$ axes
- DAt an angle of ${90^o}$ from the $x$ and $y-$ axes
So, the finding electron in a $d _{ xy }$ orbital is maximum at an angle of $45$ degree from $x$ and $y$ axis.
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Assertion $(A)$ : $\mathrm{NH}_3$ and $\mathrm{NF}_3$ molecule have pyramidal shape with a lone pair of electrons on nitrogen atom. The resultant dipole moment of $\mathrm{NH}_3$ is greater than that of $\mathrm{NF}_3$.
Reason $(R)$ : In $\mathrm{NH}_3$, the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the $\mathrm{N}-\mathrm{H}$ bonds. $\mathrm{F}$ is the most electronegative element.
In the light of the above statements, choose the correct answer from the options given below:
$\ln k=33.24-\frac{2.0 \times 10^{4} \,K }{ T }$
The Activation energy for the reaction is given by $...\,k\,\,J mol ^{-1}$. (In Nearest integer)
(Given: $R =8.3\, J\, K ^{-1} \,mol ^{-1}$ )