$t=t_{e q}(1-\alpha) \quad \alpha \quad \frac{\alpha}{2}$
$\mathrm{P}_{\mathrm{B}}=\frac{\alpha}{\left(1+\frac{\alpha}{2}\right)} . \mathrm{P}, \mathrm{P}_{\mathrm{A}}=\frac{(1-\alpha)}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}, \mathrm{P}_{\mathrm{C}}=\frac{\frac{\alpha}{2}}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}$
$K_P=\frac{P_B \cdot P_C^{\frac{1}{2}}}{P_A}$
$=\frac{(\alpha)^{\frac{3}{2}}(P)^{\frac{1}{2}}}{(1-\alpha)(2+\alpha)^{\frac{1}{2}}}$
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$(A)$ $n=3,1=0, m=0$
$(B)$ $n=4,1=0, m=0$
$(C)$ $n =3,1=1, m =0$
$(D)$ $n=3,1=2, m=1$
$(I)\,[Kr]\,5s^1$ $(II)\, [Rn]\,5f^{14}\,6d^1\,7s^2$
$(III)\,[Ar]\,3d^{10}\,4s^2\,4p^5$ $(IV)\,[Ar]\,3d^6\,4s^2$
Consider the following Statements
$(i)\, I$ shows variable oxidation State
$(ii)\, II$ is a $d-block$ element
$(iii)$ The compound formed between $I$ and $III$ is covalent
$(iv)\,IV$ shows single oxidation state
Which Statement is True $(T)$ or False $(F)$?
$(a)\;\;60\; \mathrm{mL} \frac{\mathrm{M}}{10}\; \mathrm{HCl}+40 \;\mathrm{mL} \frac{\mathrm{M}}{10} \;\mathrm{NaOH}$
$(b)\;\;55\; \mathrm{mL} \frac{\mathrm{M}}{10}\; \mathrm{HCl}+45 \;\mathrm{mL} \frac{\mathrm{M}}{10} \;\mathrm{NaOH}$
$(c)\;\;75\; \mathrm{mL} \frac{\mathrm{M}}{5}\; \mathrm{HCl}+25 \;\mathrm{mL} \frac{\mathrm{M}}{5} \;\mathrm{NaOH}$
$(d)\;\;100\; \mathrm{mL} \frac{\mathrm{M}}{10}\; \mathrm{HCl}+100 \;\mathrm{mL} \frac{\mathrm{M}}{10} \;\mathrm{NaOH}$
$pH$ of which one of them will be equal to $1$ ?