MCQ
The solution of $N{a_2}C{O_3}$ has $pH$
- ✓Greater than $7$
- BLess than $7$
- CEqual to $ 7$
- DEqual to zero
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Reason : $PbCl_4$ is powerful oxidising agent.
| Element | $IE_1$ | $IE_2$ | $IE_3$ |
| $P$ | $495.8$ | $4562$ | $6910$ |
| $Q$ | $737.7$ | $1451$ | $7733$ |
| $R$ | $577.5$ | $1817$ | $2745$ |
Then incorrect option is

$(a)$ Electron density in the $XY$ plane in $3d_x^2 - 3d_y^2$ orbital is zero
$(b)$ Electron density in the $XY$ plane in $3d_z^2$ orbital is zero
$(c)$ $2s$ orbital has one nodal surface
$(d)$ For $2pz$ orbital $YZ$ is the nodal plane.
Specify True or False.
i.$\frac{\text{K}^+}{\text{K}}=-3.02\text{V}$ ii.$\frac{\text{Cu}^{2+}}{\text{Cu}}=+0.34\text{V}$
iii.$\frac{\text{Hg}^{2+}}{\text{Hg}}=0.92\text{V}$ iv.$\frac{\text{Cr}^{3+}}{\text{Cr}}=-0.74\text{V}$
Decreasing order of reducing power of these elements is: