Test: Nomenclature Of Coordination Compounds

10 Questions MCQ Test Chemistry for JEE | Test: Nomenclature Of Coordination Compounds

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An aldotetrose has a number of pairs of enantiomers equal to:


The correct answer is option A
The aldotetroses have 2 chiral centres and so 4 different stereoisomers are possible. There are two naturally occurring stereoisomers, the enantiomers of erythrose and therefore having the D configuration but not the L enantiomers. The ketotetrose has one chiral centre and, therefore, two possible stereoisomers: erythrulose (L- and D-form).


Ferrocyanide ion is a type of​


Ferricyanide is the anion [Fe(CN)6]3−. It is also called hexacyanoferrate(III) and in rare, but systematic nomenclature, hexacyanidoferrate(III). The most common salt of this anion is potassium ferricyanide, a red crystalline material that is used as an oxidant in organic chemistry.


The number of unpaired electrons in [V(H2O)6]3+ is:


The metal ion V3+ is in +3 oxidation state, with configuration d2, and the hybridization of metal ion orbitals for ligand bond is d2sp3, where the number of unpaired electrons in [V(H2O)6]3+ is 2.


 Valence bond theory was proposed by


In the valence bond (VB) theory, proposed in large part by the American scientists Linus Pauling and John C. Slater, bonding is accounted for in terms of hybridized orbitals of the… The basis of VB theory is the Lewis concept of the electron-pair bond.


 Which of the following statements is not correct as per the nomenclature of coordination compounds is concerned?


The correct answer is option  C
In the given options, option c and d are contradictory to each other but while writing the coordination compounds also the positive ion is written on left side that is first then the negative ion is written so during nomenclature start from left that is first positive ion is written that is the cationic part first. H2O is a neutral ligand written as aqua in nomenclature of complexes and option b is a rule .So answer is option c


Metal posses two types of valency- Primary (ionizable) and secondary (non-ionizable) is the postulate of


The correct answer is option A
Metal possesses two types of valency- Primary (ionizable) and secondary (non-ionizable) is the postulate of Werner.
Werner's theory states that-
1. Metals possess two types of valencies called primary /      ionizable and secondary / non - ionizable valency.
2. Every metal atom has a tendency to satisfy both its primary and secondary valencies.
3. The ligands satisfying secondary valencies are always directed towards fixed positions in space thereby giving a definite geometry to the complex but primary valencies are non - directional. 
(a) The central metal atom\ ion in a coordination compound exhibits two types of valencies - primary and secondary

(b) Primary valencies are ionisable and correspond to the number of charges on the complex ion. Primary valencies apply equally well to simple salts and to complexes and are satisfied by negative ions.

(c) Secondary valencies correspond to the valencies that a metal atom (or) ion exercises towards neutral molecules (or) negative ions in the formation of its complex ions.

d) Secondary valencies are directional and so a complex has a particular shape. The number and arrangement of ligands in space determines the stereochemistry of a complex.


Metal-ligand bond arises by the donation of pairs of electrons by ligands to the metal is an assumption of


The correct answer is Option C.

Valence Bond Theory: 
 Assumptions and Features:-
1. Coordination compounds contain metal ions in which ligands form covalent-coordinate bonds to the metal.
2. Ligands must have a lone pair of electrons.
3. Metal should have an empty orbital of suitable energy available for bonding.
4. Atomic orbitals are used for bonding (rather than Molecular Orbitals)
5. This theory is useful to predict the shape and stability of the metal complexes.
1) Does not explain why some complexes are coloured and others are not.
2) Does not explain the temperature dependence of the magnetic properties


 Among the following which one has the highest paramagnetism?


The correct answer is Option B.
Larger the number of unpaired electrons, the more paramagnetism it has. The correct option is [Fe(H2O)6]2+
Paramagnetic character ∝ Number of unpaired electrons. So, in this paramagnetic character is maximum in [Fe(H2O)6]2+


The complex, [Ni(CN)4]2-, has:


[Ni(CN)4]2− is a square planar geometry formed by dsp2 hybridisation. [Ni(CN)4]2− is diamagnetic, so Ni2+ ion has 38 outer configurations with two unpaired electrons. 
For the formation of the square planar structure by dsp2 hybridisation, two unpaired d-electrons are paired up due to energy made available by the approach of ligands, making one of the 3d orbitals empty.


[Ni(CN)4]2- is a type of square planar complex where the oxidation state of Ni is:


Let oxidation state of Ni be a.
Therefore, a + (O.S. of CN) x4 = -2
a + (-1 x 4) = -2
a = -2 + 4
a = +2