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Correct answer is (a) rises

Average internal temperature of human body is 37°C (98.6°F)

Correct answer is (a) expands

To measure the temperature is laboratories for scientific research.

Mostly Alcohol and Mercury are used in thermometers as they remain in liquid form even with a change of temperature in them. A small change in the temperature causes change in volume of a liquid. We measure this temperature by measuring expansion of a liquid in thermometer.

1. kink 

2. 94° F,108° F 

3. 35° C or 94° F 

4. 178.45K 

5. 392.85K 

6. 10³² K

1. clinical 

2. liquid 

3. hotter object, colder object 

4. less 

5. mercury

Correct answer is (b) 37°C

(c) Expands uniformly

Correct option: (D) V²⁺

i. In case of Pt(II) complexes, the central metal atom has d8 configuration and it shows coordination number 4.

ii. According to valence bond theory, central ion undergoes dsp² hybridisation which results in formation of square planar complexes.

iii. In the Pt (IV) complexes, the central metal atom has d⁶ configuration and it shows coordination number 6. 

iv. According to valence bond theory, central ion undergoes d² sp³ or sp³ d² hybridisation which results in the formation of octahedral complexes.

All the transition metals have a tendency to form complexes. 

The tendency arises due to the following reasons:

a. They have small ionic radii. 

b. They have high effective ionic charge. 

c. Hence, they have high ratio of ionic charge to ionic radius. 

d. Transition metals and ions have vacant d-orbitals which can accomodate the lone pairs of electrons from the ligands to form coordination compounds. 

e. Transition metals show variable oxidation states. 

f. After accepting the electrons from the ligands, metal ions acquire a stable electronic configuration of the nearest inert element and form stable complexes.

eg. [Cu(NH₃)₄] ²+, [Co(NH₃)₆]³⁺, [Ni(CN)₄]^2- , etc.

ii. The stability of these complexes depends upon the nature of the metal ion, ligands and their bonding.

i. Alloys are formed by metals whose atomic radii differ by not more than 15% so that the atoms of one metal can easily take up the positions in crystal lattice of the other. 

ii. The transition metals have similar atomic radii and other characteristics, hence they form alloys very readily. 

iii. Alloys are generally solid solution which are formed by the cooling of the molten state solution of two or more transition metals. 

iv. Alloys are generally harder, have high melting points and more resistant to corrosion than the individual metals. 

v. The metals chromium, vanadium, molybdenum, tungsten and manganese are used in the formation of alloy steels and stainless steels. Ferrous alloys are the most common alloys. 

vi. Some alloys of transition metals with non-transition metals are also very common. 

eg. Brass (Cu + Zn) and Bronze (Cu + Sn)

Correct option: (C) of its tendency to attain noble gas configuration of xenon

Correct option: (D) 21 to 30

i. Electronic configurations of Sc³⁺, Ti⁴⁺, V⁵⁺ are:

Sc³⁺: [Ar] 3d⁰ ; Ti⁴⁺: [Ar] 3d⁰ ; V⁵+: [Ar] 3d⁰

ii. The ions Sc³⁺, Ti⁴⁺ and V⁵⁺ have completely empty d-orbitals i.e., no unpaired electrons are present. Thus, their salts are colourless, as d-d transitions are not possible.

Correct option: (C) only metals 

i. The electronic configuration of Cu: [Ar] 3d¹⁰ 4s¹ and Cu²⁺: [Ar] 3d⁹ . 

In copper (II) compounds, Cu²⁺ ions have incompletely filled 3d-orbital (3d⁹).

ii. The presence of one unpaired electron in 3d-orbital results in d-d transition due to which, Cu²⁺ ions absorb red light from visible spectrum and emit blue light. Therefore, copper (II) compounds are coloured.

iii. In case of zinc, the electronic configuration is Zn: [Ar] 3d¹⁰4s² and Zn²⁺: [Ar] 3d¹⁰. 

iv. Since 3d subshell is completely filled and there are no unpaired electrons, d-d transition is not possible and hence, Zn²⁺ ions do not absorb radiation in visible region. Therefore, the compounds of zinc are colourless.

The electronic configuration of Mn(II) ion (atomic number of Mn = 25) is [Ar] 3d⁵ 4s⁰ . 

It has five unpaired electrons in its d-orbitals which is a maximum value among the bivalent ions of the elements of 3d transition series. As the paramagnetic character increases with increase in the number of unpaired electrons, Mn(II) ion shows maximum paramagnetic character

Correct option: (A) Cu⁺ 

i. Transition metal ions exhibit colour due to the presence of unpaired electrons in (n-1)d-orbitals which undergo d-d transition.

ii. The metal ions which do not have unpaired electrons i.e., (n-1)d⁰ or which have completely filled d-orbitals i.e., (n-1)d¹⁰ do not absorb radiations in visible region, since d-d transitions are not possible. Hence, they are colourless ions.

eg. Cu⁺ (3d¹⁰), Ag⁺ (4d¹⁰), Zn²⁺ (3d¹⁰), Cd²⁺ (4d¹⁰), Hg²⁺ (5d¹⁰), etc.

Correct option: (B) Fe

1.Transition metals have smaller atomic size and stronger interparticle attractive forces (metallic bond) than alkaline earth metals and hence have higher density than alkaline earth metals.

2. Transition metals due to their smaller atomic size and greater effective nuclear charge, have higher ionisation energy than alkaline earth metals and hence are less electropositive.

3. Transition elements have very strong interatomic bonds due to their small size and also due to the presence of large number of unpaired electrons in their atoms. As a result, they have very high enthalpies of atomisation.

• In Ti³⁺, V²⁺, Mn²⁺, Ca²⁺d-d transition is possible. Hence they have colour.
• In Sc³⁺ and Ti⁴⁺ (3d⁰), Zn²+ and Cu+(3d¹⁰) d-d transition is not possible. 

Except Sc, the most common oxidation state of the first row transition elements is +2 which arises from the loss of two 4s electrons. The +2 state becomes more and more stable in the first half of first row transition elements with increasing atomic number because 3d orbitals acquire only one electron in each of the five 3d orbitals (i.e., each d-orbital remains half-filled) and inter electronic repulsion is the least and nuclear charge increases.

In second half of first row transition elements, electron starts pairing up in 3d orbitals and hence there is inter electronic repulsion. (Ti²⁺ to Mn²⁺ electronic configuration changes from 3d² to 3d⁵ but in second half i.e., Fe²⁺ to Zn²⁺ it changes from 3d⁶ to 3d¹⁰.)

Correct option: (A) Colourless compounds of transition elements are paramagnetic.

Correct option: (A) La, Gd and Lu

(a). A - K₂MnO₄ 

B - KMnO₄

2MnO₄ + 4KOH + O₂ \(\longrightarrow\) 4K₂MnO + 2H₂O

K₂MnO₄ \(\overset{H_2SO_4}\longrightarrow\)  KMnO₄

(b).  10I^- +2MnO₄^- + 16H⁺ \(\longrightarrow\) 2Mn²⁺ + 8H₂O +5I₂

5Fe²⁺ + MnO₄- + 8H⁺ \(\longrightarrow\) Mn²⁺ + 4H₂O + 5Fe³⁺

(c) Decrease in ionic radii.

Correct option: (C) hydrogenation

1. Yes 

2. Transition elements form complex compounds due to the comparatively smaller size of the metal ion, their high ionic charges and the availability of d-orbital for the bond formation.

3. [Fe(CN)₆]^3- , [Cr(NH₃)₆]³⁺

Correct option: (D) have variable valency

Correct option: (B) +7

Correct option: (A) contract slightly

(1). On heating, KMnO₄ decomposes at 513 K into potassium manganate and oxygen gas is evolved.

2KMnO₄ \(\overset{Heat}\longrightarrow\)  K₂MnO₄ + MnO₂ + O₂

(2). Two examples to show the oxidising property of KMnO₄

In acid medium it oxidises iodide to iodine.

10I^- + 2MnO₄^- + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂ 

In neutral or faintly alkaline solutions it oxidises iodide to iodate.

2MnO₄^- + H₂O + I^- → 2MnO₂ + 2OH^- + IO₃^-

(iii) Hydration enthalpy

Correct option: (C) third group and 7^th and 6^th period of periodic table

Actinoid contraction is greater than lanthanoid contraction:

i. The size of the atoms or ions of actinoids decrease regularly along the series with the increase in atomic number from actinium to lawrencium. This steady decrease in the ionic radii with the increase in atomic number is called actinoid contraction.

ii. The actinoid contraction is due to the imperfect shielding of 5f-electron. 

Despite of the imperfect shielding of 5f-orbitals, the effective nuclear charge increases which results in contraction of the size.

iii. It may be noted that in actinoid contraction, there are bigger jumps in ionic size between the consecutive members as compared to lanthanoids.

iv. This is due to lesser shielding of 5f-electrons (as compared to shielding of 4f-electrons in lanthanoids) which results in greater increase in the effective nuclear charge and therefore, larger attraction.

Electronic configuration of Th⁴⁺ = [Rn] 5f⁰ 6d⁰ 7s⁰

Correct option: (A) U

i. Actinoids have variable oxidation states ranging from +2 to +7 due to availability of 5f, 6d and 7s orbitals.

ii. The common oxidation state of actinoid elements is +3. +3 oxidation state is formed by loss of two 7s and one 5f or 6d electrons.

iii. With increase in atomic number, +3 oxidation state becomes more and more stable.

iv. Beside +3 oxidation state, actinoids show +2, +4, +5, +6 and +7 oxidation states.

a. Elements Am and Th show +2 oxidation state. 

eg. ThI₂, ThS, ThBr₂, etc. 

b. Elements Th, Pa, U, Np, Pu, Am and Cm show +4 oxidation state. 

c. Elements Th, Pa, U, Np, Pu and Am also show +5 oxidation state. 

d. Elements U, Np, Pu and Am show +6 oxidation state.

Note: When the oxidation number increases to +6, the actinoid ions form oxygenated ions due to high charge density.

eg. UO₂²⁺, NpO₂²⁺ ,  etc.

e. Np and Pu show +7 oxidation states.

v. Actinoids exhibit large number of variable oxidation states because all the electrons in 5f, 6d and 7s orbitals can take part in bond formation due to very small energy gap between these orbitals.

vi. Actinoids have more compounds in +3 oxidation state than in +4 oxidation state. However, compounds of actinoids in +3 and +4 oxidation states have tendency to undergo hydrolysis.

Note: Oxidation states of actinium and actinoids:

Because they have completely filled d-orbitals in their atomic as well as stable ionic state.

In actinoids, 5f and 6d subshells are close in energy. The outermost 7s orbital remains filled with 2 electrons (7s²). The electron can easily jump from 5f to 6d or vice versa. Further, irregularities in electronic configurations are also related to the stabilities of f⁰ , f⁷ and f¹⁴ occupancy of the 5f-orbitals. Hence, they show a large number of oxidation states (Moreover, they are radioactive with short half-lives. Hence, their properties cannot be studied easily).

i. Pm³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰4s² 4p⁶ 4d¹⁰5s² 5p⁶ 4f⁴ .

ii. Ce⁴⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰4s² 4p⁶ 4d¹⁰5s² 5p⁶ .

iii. Lu²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰4s² 4p⁶ 4d¹⁰5s² 5p⁶ 4f¹⁴5d¹ .

Alec Jeffreys.

In the lanthanoids, 4f and 5d subshells are very close in energy. The outermost 6s-orbital remains filled with two electrons (6s²). The electrons can easily jump from 4f to 5d or vice-versa. Further, irregularities in electronic configurations are also related to the stabilities of f⁰ , f⁷ and f¹⁴ occupancy of f-orbitals. Hence, their electronic configurations are not known with certainty.

Electronic configuration of lanthanum and 4f-series of f-block elements:

i. The 4f-series includes elements from cerium (Ce) to lutetium (Lu). The electronic configuration of these elements can be expressed in terms of its nearest inert gas Xe (Z = 54).

ii. Electronic configuration of Xe (Z = 54) = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹0 5s² 5p⁶ Therefore, general electronic configuration of 4f-series is [Xe] 4f^1-14 5d^0-1 6s² .

iii. Lanthanum has electronic configuration [Xe]4f⁰ 5d¹ 6s² . It does not have any 4f electrons.

(1). Yes.

(2). It is due to Lanthanoid contraction. In Lanthanoid series the 4f electrons are being added in the antipenultimate shell. The 4f electrons have very poor shielding effect because of diffused shape of 4f orbitals and the effective nuclear charge experienced by each 4f electrons increases. Hence there is a regular decrease in radii with increase in atomic number.

3. Consequences of lanthanoid contraction are :

• Difficulty in separation of lanthanoids due to similarity in chemical properties.
• Similarity in size of elements belonging to same group of second & third transition series. 

i. It is observed that lanthanum, gadolinium and lutetium show different electronic configuration because the 5d and 4f-orbitals are nearly of the same energy and the distinction between the two is difficult.

ii. Due to this, some extra stability is achieved when the 4f is half-filled and completely filled so the next electron goes in 5d-orbital instead of 4f-orbital. 5d orbital contains one electron in 5d subshell in case of La, Gd and Lu while it is empty in case of other lanthanoids.

iii. f⁰ , f⁷ and f¹⁴ configurations have extra stability due to empty, half filled and completely filled f orbitals respectively.

iv. The electronic configuration of ₅₇La, ₆₄Gd and ₇₁Lu in +3 oxidation state are as follows:

La⁺³ : [Xe] 4f⁰ ; 

Gd⁺³ : [Xe] 4f⁷ ; 

Lu⁺³ : [Xe] 4f¹⁴.

Correct answer is 38.4%

a) K₂Cr₂O₇ is prepared from chromate one Fe Cr₂ O₄. 

Step I: The powdered ore is heated with molten alkali in free access of air to form soluble sodium chromate.

4 Fe Cr₂ O₄ + 16 NaOH + 7O₂ → 8 Na₂CrO₄ + 2 Fe₂O₃ + 8 H₂O

Step II: Sodium chromate (Na₂Cr O₄) is filtered and acidified with dil. H₂SO₄ to form sodium dichromate.

2 Na₂CnO₄ + H₂SO₄ → Na₂Cn₂O₇ + Na₂SO₄ + H₂O

Step III: Na₂Cn₂O₇ solution is treated with KCI to form K₂Cr₂O₇.

Na₂Cn₂O₇ + 2 KCI → K₂Cn₂O₇ + 2NaCI

b) Consequences of lanthanoid contraction are

• Difficulty in separation of lanthanoids due to similanity in chemical properties.
• The similarity in size of elements belonging to same group of second & third transition series.