MCQ
The element having electronic configuration belongs to $n{s^2}(n - 1){d^{1 - 10}}(n - 2){f^{1 - 14}}$
- A$s-$ block
- B$p-$ block
- C$d-$ block
- ✓$f-$ block
✓
Answer
Correct option: D.
$f-$ block
d
The general electronic configuration of $f - b$ lock is $ns ^2( n -1) d ^{1-10}( n -2) f ^{1-14}$.
The general electronic configuration of $f - b$ lock is $ns ^2( n -1) d ^{1-10}( n -2) f ^{1-14}$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If 22 gm of benzene $\left( C _6 H _6\right)$ is dissolved in 122 gm of carbon tetrachloride $\left( CCl _4\right)$ calculate the mass percentage of benzene.
→${N_2}$ forms $NC{l_3}$, whereas $P$ can form both $PC{l_3}$ and $PC{l_5}$. Why
→In the following reaction, $Image$ the structure of the major product ' $X$ ' is
→In the reaction
→$C{{H}_{3}}CN+2H\underset{\text{Ether}}{\mathop{\xrightarrow{HCl}}}\,X\underset{{{H}_{2}}O}{\mathop{\xrightarrow{\text{Boiling}}}}\,Y;$ the term $Y$ is:
The experimental data for the reaction $2A + {B_2} \to 2AB$ isThe rate equation for the above data is
→|
Exp. |
$[A]_0$ |
$[B]_0$ |
Rate (mole $s^{-1}$) |
|
$(1)$ |
$0.50$ |
$0.50$ |
$1.6 \times {10^{ - 4}}$ |
|
$(2)$ |
$0.50$ |
$1.00$ |
$3.2 \times {10^{ - 4}}$ |
|
$(3)$ |
$1.00$ |
$1.00$ |
$3.2 \times {10^4}$ |
$C{H_3}COOH\mathop {\xrightarrow{\Delta }}\limits_{{P_2}{O_5}} X$ . Identify $X$
→The magnetic moment of $[MnX_4]^{2-}$ is $5.9\, BM.$ The geometry of the complex ion is : ($X =$ monodentate halide ion)
→Assertion : The water pouch of instant cold pack for treating athletic injuries breaks when squeezed and $NH_4NO_3$ dissolves lowering the temperature.
→Reason : Addition of non-volatile solute into solvent results into depression of freezing point of the solvent.
Which one of the following statements is true for $HN{O_2}$
→Nitrogen gas is liberated by thermal decomposition of