Solution

[Pt(en)₂Cl₂] - Dichloridobis(ethane-1,2-diammine)platinum (II)
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or
(i) [Co(NH₃)₆]₂(SO₄)₃

(ii) K₃[Cr(C₂O₄)₃]

Solution

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(i) [CoF₆]³⁻ is paramagnetic due to presence of four unpaired electrons.
(ii) [Co(en)₃]³⁺ is more stable due to chelation.
(iii) [Co(en)₃]³⁺ forms inner orbital complex (d²sp³).
(iv) [CoF₆]³⁻ forms high spin complex (sp³d²).

Solution

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Solution

Solution

Rate law for a zero-order reaction can be given as k = ([R]⁰ – [R] )/ t
Substituting the given values-
4 × 10⁻³ Ms⁻¹ = (0.1 – 0.064) M / t
t = 0.036 M/4 × 10⁻³ Ms⁻¹ = 9 s
So, the time taken would be 9 seconds.

Solution

(i) Activated charcoal present in gas mask adsorbs the harmful suffocating gases which surround the user and protects them from the particulate matter and ashes present in coal mines.
(ii) When FeCl₃ is added to NaOH solution, a negatively charged sol is obtained with adsorption of OH⁻ ions. Sol is represented as Fe₂O₃.xH₂O/OH⁻.
(iii) With increase in temperature chemisorption first increases as molecules get activation energy for adsorption. After certain temperature, chemisorption decreases.
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Solution

volume of 108 g of elements= mass/density=10 cm³
Volume of one unit cell =(300 pm)³= 2.7×10⁻²³cm³
Numbers of unit cell in 10 cm³=10/2.7×10⁻²³ cm³=10²⁴/2.7 unit cells
one unit cell have 4 atoms so total numbers of atoms =10²⁴/2.7 unit cells×4=1.48×10²⁴ atoms.

Solution

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Solution

(a) (i) Zone refining
(ii) Distillation
(b) Chemical reactions involved in the extraction of Cu from Cu2S are:
Cu₂S ore is heated in a reverberatory furnace after mixing with silica. In the furnace, iron oxide slags of as iron silicate and copper are produced in the form of copper matte. This contains Cu2S and FeS.
The reactions undergo as given below.
A. 2FeS + 3O2 → 2FeO + 2SO₂
B. FeO + SiO₂ → FeSiO₃
C. 2Cu₂S + 3O2 → 2Cu₂O + 2SO₂
D. 2Cu₂O + Cu₂S → 6Cu + SO₂

Solution

(i) Transition elements and their compounds act as a catalyst due to their ability to adopt multiple oxidation states and to form complexes.

(ii) Mn²⁺ has d⁵ configuration (stable half-filled configuration) and it prefers to stay in +2 oxidation state, so, E⁰ value for Mn²⁺/ Mn is negative. Whereas the high energy to transform Cu(s) to Cu²⁺ (aq) is not balanced by its hydration enthalpy, hence Cu2+ is not favoured over Cu state and it has positive E⁰ value.

(iii) Due to extra stability of empty, half-filled and fully filled/subshells, Actinoids show irregularities in their oxidation states.