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a year agoContributor-Level 10
8.41 The atomic number of Cerium (Ce) is Z = 58.
The electronic configuration of 58Ce= [Xe]54 4f1 5d1 6s2
And, the electronic configuration of Ce3+= [Xe]54 4f1 , i.e., there is only one unpaired electron, i.e., n = 1.
Now, the magnetic moment on the basis of spin only formula is given as:
μ = √n(n+2) BM
⇒ μ = √1(1+2)
⇒ μ = √3 BM
⇒ μ = 1.73 BM
Bohr magneton or BM is a unit for expressing the magnetic moment of an electron caused by its orbital or spin angular momentum.
New answer posted
a year agoContributor-Level 10
8.40 Last actinoid is lawrencium - 103
Electronic configuration: [Rn]86 5f14 6d1 7s2
Possible oxidation state: +3
New answer posted
a year agoContributor-Level 10
8.39 Lanthanoids show oxidation states of +2, +3, +4 because of the large energy gap between 5d and 4f sub-shells while actinoids show oxidation states of +3 to +7 because of the small energy difference between 5f, 6d, and 7s orbitals.
New answer posted
a year agoContributor-Level 10
8.38 The inner transition elements are those in which the last electron enters the f-subshell. The lanthanoids atomic number 58-71 and actinoids 90-103. The atomic numbers 59, 95 and 102 are inner transition elements
New answer posted
a year agoContributor-Level 10
8.37 An alloy is a homogenous mixture of two or more metals or non-metals. Lanthanoid is used for the production of alloy steels for plates and pipes called as mischmetals. It contains about 95% lanthanoid metal, iron and traces of Al, S, C etc. in traces. It is used in magnesium- based alloy to produce bullets, shells and lighter flint.
New answer posted
a year agoContributor-Level 10
8.36 (i) The chromite ore (FeCr2O4) is fused with sodium hydroxide (NaOH)
4FeCr2O4 +16 NaOH+ 7O2 → 8Na2CrO4 +2Fe2O3 +8H2O
The yellow solution of sodium chromate is then filtered and acidified with sulphuric acid giving its dichromate.
2Na2CrO4 +H2SO4→ Na2Cr2O7 + Na2SO4 +H2O
On cooling, sodium sulphate crystallizes out as Na2SO4.10H2O and is removed.
Na2Cr2O7 +2KCl → K2Cr2O7 + 2NaCl
(ii) The pyrolusite ore(MnO2) is oxidised in the presence of potassium hydroxide by heating.
2MnO2 + 4KOH +O2→ 2K2MnO4 + 2H2O
The green potassium manganate (K2MnO4) is then treated with a current of chlorine or ozone to oxidise potassium manganate to potassi
New answer posted
a year agoContributor-Level 10
8.35 (i) The lower oxide have low oxidation state while the higher oxide has high oxidation state, example MnO is basic and Mn2O7 is
(ii) Oxygen and fluorine have a small size and high electronegativity and can easily oxidize metals, example V2O5.
(iii) Oxoanions of metals have higher oxidation states because of high electronegativity of oxygen and highly oxidizing property example, Cr in CrO7 2- has an oxidation state of +6
New answer posted
a year agoContributor-Level 10
8.34 From the table given below:
Mn3+ : 3d4 | unpaired electrons = 4 |
V3+ : 3d2 | unpaired electrons =2 |
Cr3+ : 3d3 | unpaired electrons= 3 |
Ti3+ : 3d1 | unpaired electrons =1 |
Cr3+ is most stable in aqueous solution because of half filled d-orbital.
New answer posted
a year agoContributor-Level 10
8.33 Copper (29) has electronic configuration 1s22s22p63s23p63d104s1. It can easily lose one electron to give stable configuration as it has completely filled d-orbital.
New answer posted
a year agoContributor-Level 10
8.32 The reactions in which same substance gets oxidized as well as reduced due to unstable oxidation state.
Example:
2MnO42- + 4H+? 2MnO4- + MnO2 + 2H2O
Mn (VI) is oxidised to Mn (VII) and also reduced to Mn (IV).
2CrO4 3- +2H+? CrO4 2- +Cr3+ + 4H2O
Cr (V) is oxidised to Cr (VI) and also reduced to Cr (III).
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