Metallic character decreases
Non-metallic character increases
$\Rightarrow$ It is due to increase in ionization enthalpy and increase in electron gain enthalpy.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

The $E _{\text {ceil }}$ for the given cell is $0.1115\,V$ at $298\,K$ when $\frac{\left[ M ^{+}( aq )\right]}{\left[ M ^{3+}( aq )\right]}=10^{ a }$
The value of a is
Given : $E _{ M ^{3+} / M ^{+}}=0.2\,V$
$\frac{2.303 RT }{ F }=0.059\,V$
$A.$ For 1 s orbital, the probability density is maximum at the nucleus.
$B.$ For $2 s$ orbital, the probability density first increases to maximum and then decreases sharply to zero.
$C.$ Boundary surface diagrams of the orbitals encloses a region of $100 \%$ probability of finding the electron.
$D.$ $p$ and d-orbitals have $1$ and $2$ angular nodes respectively.
$E.$ Probability density of p-orbital is zero at the nucleus.