NEET Practice Questions (MCQs) with Answers & Solutions

$\left[\mathrm{Mn}\right(\mathrm{CO}{)}_4\mathrm{NO}]$ is diamagnetic because:

If CO ligands are substituted by NO in respective neutral carbonyl compounds then which of the following will not be correct formula ?

Which of the following species can act as reducing agent ?

What is electronic arranegment of metal atom/ionin octahedral complex with ${\mathrm{d}}^4$ configuration , if ${∆}_0$< pairing energy ?

Which of the following statement is not correct ?

Give the correct of initial T or F for following statements. Use T is statement is true and F if it is false

(I) Co(III) is stabilised in presence of weak ligands , while Co(II) is stabilised in presence of strong field ligand.

(II) Four coordinated complexes of Pd(II) and Pt(II) are diamagnetic and square planar.

(III) $\left[\mathrm{Ni}\right(\mathrm{CN}{)}_4{]}^{4-}$ ion and $\left[\mathrm{Ni}\right(\mathrm{CO}{)}_4]$ are diamagnetic tetrahedral and square planar.

(IV) ${\mathrm{Ni}}^{2+}$ ion does not inner orbital octahedral complexes.

Match List-I with List-II and select the correct answer using the codes given below :

                      List-I                                              List-II

(I) $[{\mathrm{FeF}}_6{]}^{3-}$                                                   (A) 1.73 BM

(II) $\left[\mathrm{Ti}\right({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$                                             (B) 5.93 BM

(III) $\left[\mathrm{Cr}\right({\mathrm{NH}}_3{)}_6{]}^{3+}$                                            (C) 0.00 BM

(IV) $\left[\mathrm{Ni}\right({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$                                             (D) 2.83 BM

(V)  $\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6{]}^{4-}$                                               (E) 3.88 BM

          (I)    (II)   (III)   (IV)   (V)                       (I)    (II)   (III)   (IV)   (V) 

Set of d-orbitals which is used by central metal during formation of ${\mathrm{MnO}}_4^{-}$?

${\mathrm{FeSO}}_4$ is a very good absorber for NO, the new compound formed by this process is found to contain number of unpaired electrons:

$\mathrm{A}\left[\mathrm{M}\right({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$ complex typically absorbs at around 600 nm. It is allowed to react with ammonia to form a new complex $\left[\mathrm{M}\right({\mathrm{NH}}_3{)}_6{]}^{2+}$ that should have absorption at :