Download App

4. d and f- block elements — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistry4. d and f- block elements1 Mark
MCQ
The electronic configuration for Neodymium is: [Atomic Number for Neodymium $60$]
  • ✓
    $[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$
  • B
     $[\mathrm{Xe}] 5 \mathrm{f}^4 7 \mathrm{~s}^2$
  • C
     $[\mathrm{Xe}] 4 \mathrm{f}^6 6 \mathrm{~s}^2$
  • D
     $[\mathrm{Xe}] 4 \mathrm{f}^1 5 \mathrm{~d}^1 6 \mathrm{~s}^2$
✓

Answer

Correct option: A.
$[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$
a
Electronic configuration of $\mathrm{Nd}(\mathrm{Z}=60)$ is; $[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 4. d and f- block elements→4. d and f- block elements — all question types→Chemistry STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Consider the following single step reaction in gas phase at constant temperature.

$2 \mathrm{~A}_{(\mathrm{g})}+\mathrm{B}_{(\mathrm{g})} \rightarrow \mathrm{C}_{(\mathrm{g})}$

The initial rate of the reaction is recorded as $r_1$ when the reaction starts with $1.5 \mathrm{~atm}$ pressure of $\mathrm{A}$ and $0.7 \mathrm{~atm}$ pressure of B. After some time, the rate $r_2$ is recorded when the pressure of $C$ becomes $0.5 \mathrm{~atm}$. The ratio $r_1: r_2$ is $\qquad$ $\times 10^{-1}$.

(Nearest integer)

→
Benzophenone can be converted into benzene using
→
The correct order of acidic strength of oxy-acids of chlorine is
→
In the reduction of nitrobenzene, which of the following is the intermediate
→
Find out the incorrect statement.

→
Given that $\frac{1}{3}\mathop {{\lambda _m}}\limits^\infty  \left( {F{e^{3 + }}} \right) = 68\,oh{m^{ - 1}}\,c{m^{ - 1}}\,e{q^{ - 1}}$ and $\frac{1}{2}\mathop {{\lambda _m}}\limits^\infty  \left( {SO_4^{2 - }} \right) = 80\,oh{m^{ - 1}}\,c{m^{ - 1}}\,e{q^{ - 1}}$ What will be value of $\mathop {{\lambda _{eq}}}\limits^\infty  \left( {F{e_2}{{\left( {S{O_4}} \right)}_3}} \right)$ ? ............ ${\rm{oh}}{{\rm{m}}^{ - 1}}{\mkern 1mu} {\rm{c}}{{\rm{m}}^2}$ $\mathrm{eq}^{-1}$
→
The correct order of acidity for the following compounds is
→
The reaction given below is known as

${C_2}{H_5}ONa + I{C_2}{H_5}\xrightarrow{{}}{C_2}{H_5}O{C_2}{H_5} + NaI$

→
${K_f}$ of $1,4\,-$ dioxane is $4.9\,mo{l^{ - 1}}$ for $1000\,g$. The depression in freezing point for a $ 0.001\,m$  solution in dioxane is
→
The compound($s$) having peroxide linkage is(are)

$(A)$ $H _2 S _2 O _7$  $(B)$ $H _2 S _2 O _8$  $(C)$ $H _2 S _2 O _5$  $(D)$ $H _2 SO _5$

→