MCQ
Electronic configuration of $Gd(64)$ is
- A$[Xe]\, 4f^2,\,5d^9,\,6s^2$
- ✓$[Xe]\, 4f^7,\,5d^1,\,6s^2$
- C$[Xe]\, 4f^6,\,5d^2,\,6s^2$
- D$[Xe]\, 4f^7,\,5d^0,\,6s^2$
✓
Answer
Correct option: B.
$[Xe]\, 4f^7,\,5d^1,\,6s^2$
b
The electronic configuration of $La ( Z =57)$ is $[ Xe ] 5 d ^1 6 s ^2$. Therefore, further addition of electrons occurs in a lower energy $4 f$-orbital till it is exactly half filled at $Eu ( Z =63)$. Thus, the electronic configuration of $Eu$ is $[ Xe ] 4 f ^7 6 s ^2$.
The electronic configuration of $La ( Z =57)$ is $[ Xe ] 5 d ^1 6 s ^2$. Therefore, further addition of electrons occurs in a lower energy $4 f$-orbital till it is exactly half filled at $Eu ( Z =63)$. Thus, the electronic configuration of $Eu$ is $[ Xe ] 4 f ^7 6 s ^2$.
Therefore, addition of next electron doesn't occur in a more stable exactly halffilled $4 f ^7$ shell but occur in a little higher energy $5 d$-orbital. Thus, the electronic configuration of $Gd ( Z =64)$ is [Xe] $4 f ^7 5 d ^1 6 s ^2$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Which of the following gives layer test ?
$EMF$ of following concentration cell will be .............. $\mathrm{V}$
→$C{l_2}\,(1\,{\mkern 1mu} atm)|\,\,(C{l^ - }(1{\mkern 1mu} M)||C{l^ - }\,(0.1{\mkern 1mu} M)|\,\,C{l_2}\,(2\,{\mkern 1mu} atm)$
The electrolytic decomposition of dilute sulphonic acid with platinum electrode in cathodic reaction is
When ${H_2}S$ gas is passed through nitric acid, the product is
→Upon mixing $50.0\,mL$ of $0.1\,M$ lead nitrate solution with $50\,mL$ of $0.05\,M$ chromic sulphate, solution, precipitation of lead sulphate solution, takes place. How many moles of, lead sulphate are, formed and what is the molar concentration, of chromic sulphate left in the solution ?
→Major product of the following reaction
→$C{H_3} - C{H_2} - Br\,\xrightarrow{{KCN}}\,\xrightarrow[{{H_2}O}]{{\bar OH}}\,\xrightarrow[\Delta ]{{PC{l_5}}}$ product
The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at $300\, K$ is $1.0 \times 10^{-3} s ^{-1}$ and the activation energy $E _{ a }=11.488\, kJ\, mol ^{-1},$ the rate constant at $200\, K$ is ............ $\quad \times 10^{-5} s ^{-1} .$
→(Round of to the Nearest Integer).
(Given : $\left. R =8.314\, J\, mol ^{-1} K ^{-1}\right)$
Order of bond length of bond $x, y, z$
→How much water is needed to dilute $10 \, ml$ of $10 \, N$ hydrochloric acid to make it exactly decinormal ($0.1\,N$) .......... $ml$
→Phosphine may be produced by adding water to