Test: Single Correct MCQs: The d- and f-Block Elements & Coordination Compounds | JEE Advanced - JEE MCQ

∵ Basicity of hydroxides decreases on moving left to right in a period.

German silver is alloy of Cu + Zn + Ni

It has 2 unpaired electrons. 3d orbital of Ni²⁺ ion. At No. of Ni = 28.

Hypo solution (Na₂S₂O₃) is used in photography to remove the unaffected AgBr in the form of soluble complex.

Conc. HNO₃ renders iron passive by forming a thin protective film of Fe₃O₄ on its surface.

Zinc-copper couple is obtained by mixing zinc dust with copper guaze.

Mn ²⁺ in Mn SO₄.4H₂O has d 5 con figur ation (five un paired electr on s); Cu²⁺ in CuSO₄.5H₂O has d 9 configuration (one unpaired electron); Fe²⁺ in FeSO₄.6H₂O has d⁶ configuration (four unpaired electron); and Ni²⁺ in NiSO₄.6H₂O has d⁸ configuration (two unpaired electron). Thus CuSO₄.5H₂O has lowest degree of paramagnetism.

Solder is an alloy containing Sn – 67% and Pb – 33%.

KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 3Cl₂

CuSO₄ + 2KCN → Cu (CN)₂ + K₂SO₄
2Cu(CN)₂ → Cu₂(CN)₂ +  (CN)₂ (Cyanogen)
Cu₂(CN)₂ + 6 KCN → 2K₃ [Cu(CN)₄]

HgS does not dissolved in hot dil. HNO₃

TIPS/FORMULAE :

Chrome alum is K₂SO₄.Cr₂(SO₄)₃.24H₂O

The filtrate is yellow due to  ion and residue is brown due to Fe(OH)₃.

Hence green coloured powder blown in the air is Cr₂O₃.

The reaction of   in acidic medium is derived as follows:

Organometallic compounds are those compounds in which metal atom is directly bonded with C-atom. H₃C – Li.

The electronic configurations of cations in the given salts are

Only Cu²⁺ ion has one unpaired electron in 3d orbital and so, its salt is expected to be coloured.

The structure of 

There are six normal Cr – O bonds and two bridged Cr – O bonds.
The six normal Cr – O bonds are expected to be equivalent and different from those of the bridged Cr – O bonds.

NOTE : In metal carbonyl the metal is in zero oxidation state.
In Ni(CO)₄ , O.N. of Ni = 0
For Ni (Z = 28)

In presence of CO two 4s electrons pair up, thus

In Ni(PPh₃)₂Cl₂, O.N. of Ni = +2 For Ni²⁺

PPh₃ and Cl^– can’t pair up d-electrons, leading to sp³ hybridization leading to tetrahedral geometry.

Haematite ore (Fe₂O₃) is first reduced to cast iron which is then oxidised for removing carbon (impurity) as CO₂.

In [MnO₄]^–, Mn is in +7 oxidation state.
Electronic configuration of Mn (Z = 25) : [Ar] 3d⁵4s²
Electronic configuration of Mn⁷⁺  : [Ar] 3d⁰4s⁰ Central atom in other ions have definite number of d electrons
No. of electrons

2Fe + 3Cl₂ (dry) → 2FeCl₃ (anhydrous)

Stable oxidation state of Mn in alkaline medium is +6.
So, MnO₂ is oxidised to K₂MnO₄ (purple green) by atmospheric oxygen in KOH medium.

The configuration of Ni²⁺ is 3d⁸. For the elements of the first transition series, Cl^– behaves as a weak field/ high spin ligand. Hence Ni in [NiCl₄]^2– is sp³  hybridised leading to tetrahedral shape.

TIPS/FORMULAE :

 (m= spin magnetic moment) Here Co is present as Co²⁺ ion which has 3 unpaired electrons. So the spin magnetic moment will be