MCQ
Magnetic moment of the complexes is zero in
- ✓$[Ni(CN)_4]^{2-} , [Ni(CO)_4]$
- B$[Ni(H_2O)_6]^{2+} , [Ni(CN)_4]^{2-}$
- C$[Ni(H_2O)_6]^{2+} , [Ni(CO)_4]$
- D$[Fe(CN)_6]^{4-} , [Fe(CN)_6]^{3-}$
✓
Answer
Correct option: A.
$[Ni(CN)_4]^{2-} , [Ni(CO)_4]$
a
$\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2},\,\,\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ no unpaired $e^-$
$\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2},\,\,\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ no unpaired $e^-$
$\mu_{\mathrm{s}}=0$
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
The element which never acts as reducing agent in a chemical reaction is
In the electrochemical cell :
→$Zn \,|\,ZnSO_4\,(0.01\,M)\,||\,CuSO_4\,(1.0\, M)\,|\,Cu,$
the $emf$ of this Daniell cell is $E_1.$ When the concentration of $ZnSO_4$ is changed to $1.0\, M$ and that of $CuSO_4$ changed to $0.01\, M,$ the $emf$ changes to $E_2.$ Fromthe followings, which one is the relationship between $E_1$ and $E_2$ ? (Given, $RT/F = 0.059$)
Super are derived from compounds of
Where $Nu =$ Nucleophile
→Find out the correct statement from the options given below for the above $2$ reactions.
$\underset{{{C}_{5}}{{H}_{10}}O}{\mathop{(A)}}\,\xrightarrow{{{H}_{3}}{{O}^{\oplus }}}B+C;\,(B)$ and $(C)$ both give $+ve$ iodoform test. Compound $(A)$ is
→(Consider all steps and intermediate) correct statement is
Which of the following does not have valence electron in $3d-$ subshell
→The addition of metals like $Cr,\,Mn,\,W$ and $Ni$ to ordinary steel makes it
→$3{d^0}4{s^0}$ electronic configuration exhibits
→Homoleptic complex from the following complexes is