1 Marks Question

Question 11 Mark

'Spin only' magnetic for $Cu _{(\rho q )}^{+2}$ ion. Calculate the magnetic moment.

Answer

$\mu=\sqrt{n(n+2)}$ B.M. $=\sqrt{1(1+2)}=\sqrt{3}$ B.M. $=1.7 B . M$.

MCQ 21 Mark

Give reasons why $Zn , Cd , Hg$ and Cn are not transition elements.

Answer

The reason why $Zn , Cd , Hg$ and Cn are not considered transition elements is because of their fully filled d-orbitals.

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Question 31 Mark

Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Answer

Following are the oxo-metal ions of the first transition series in which the metal exhibit an oxidation state equal to the group number of the transition series.
Vanadate $VO _3^{-}, V (+5) \quad$ 5th group
Chromate $CrO _4^{2-}, Cr (+6) \quad$ 6th group
Permanganate $MnO _4^{-}, Mn (+7) \quad$ 7th group

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Question 41 Mark

Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

Answer

Atomic Number (Element)	Electronic Configuration
61 (Pm)	$[ Xe ] 4 f^5 6 s^2$
91 (Pa)	$[ Rn ] 5 f^2 6 d^1 7 s^2$
101 (Md)	$[ Rn ] 5 d^{13} 7 s^2$
109 (Unu)	$[ Rn ] 5 f^{14} 6 d^7 7 s^2$

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Question 51 Mark

Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Answer

The last element of the actinoid series is Lawrencium. Its electronic configuration is as follows :
${ }_{103} Lr =[ Rn ] 5 f^{14} 6 d^1 7 s^2$
The possible oxidation state of Lr is +3 because it has all round stable configuration (4f¹⁴).
$Lr ^{3+}=[ Rn ] 4 f ^{14}$

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Question 61 Mark

What are inner transition elements? Decide which of the following atomic numbers are the atomic number of the inner transition elements :
29, 59, 74, 95, 102, 104.

Answer

Inner transition elements are those elements in which the outermost three shells of the atoms or in any oxidation state are incomplete and their f orbitals are incomplete. These are the elements of f section. They have two series : (i) Lanthanoids (Z = 58 to 71) and (ii) Actinoids (Z = 90 to 103).
Of the above, atomic numbers 59, 95 and 102 belong to internal transition elements.

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Question 71 Mark

What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Answer

Alloy: A homogeneous mixture of two or more metals or a metal and a non-metal is called alloy.
A major alloy containing lanthanoid metals is Misch metal, which contains ~95% a lanthanouid metal, ~5% iron and small amounts of SC, Ca and Al. Misch metal is found in large quantities in magnesium based alloys used in the production of gun bullets, armour, and light flint.

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Question 81 Mark

Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

Answer

In the post first transition series, Cu most often shows the stable oxidation state (+1), because this results in the 3d¹⁰ stable configuration, Cu⁺¹ = [Ar] 3d¹⁰.

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Question 91 Mark

Predict which of the following will be coloured in aqueous solution? $T i ^{3+}, V ^{3+}, C u ^{+}, S c ^{3+}$,$Mn ^{2+}, Fe ^{3+}$ and $Co ^{2+}$. Give reasons for each.

Answer

All ions except Sc³⁺ are coloured in aqeuous solution because they contain unpaired electrons in d-orbitals, hence d-d transition occurs easily in them which is responsible for their colour.

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Question 101 Mark

How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.

Answer

There is a difference of one in the oxidation states of post-transition elements. For example. Manganese shows +2, +3, +4, +5, +6, +7 states, whereas non-transition elements like p-block elements always have a difference of two, such as +2, +4 or +3, +5 or +4, +6 etc.

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Question 111 Mark

Explain giving reason :
Transition metals and their many compounds act as good catalyst.

Answer

Transition metals and many of their compounds are good catalyst because they have variable valency (oxidation number) and the property of forming complex compounds in which unpaired electrons are used.

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Question 121 Mark

Explain giving reason :
The transition metals generally form coloured compounds.

Answer

Compounds of transition metals are generally coloured because they contain unpaired electrons which allow d-d transition $\left(t_{2 g}\right.$ to $\left.e_g\right)$ to occur easily by visible light. In the presence of ligand (water etc.) the d orbital splits into two parts $-t_{2 g}$ and e_g. For this reason, their colour is observed only in aqueous solution or hydrated state.

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Question 131 Mark

Explain giving reason :
Transition metals and many of their compounds show paramagnetic behaviour.

Answer

Transition metals and most of their compounds are paramagnetic, because in them unpaired electrons are found near the metal and those elements or compounds which have unpaired electrons are paramagnetic. Like Sc = [Ar] 3d¹ 4s² it has one unpaired electron hence it is paramagnetic, similarly in FeSO₄, Fe⁺²has four unpaired electrons so it is also paramagnetic.

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Question 141 Mark

What are the different oxidation states exhibited by the lanthanoids?

Answer

The general oxidation state of lanthanoids is +3 but some lanthanoids also exhibit +2 and +4 oxidation states. Like Eu²⁺ and Ce⁺⁴.

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Question 151 Mark

Write the names of the components of alloy brass.

Answer

Copper, Zinc.

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Question 161 Mark

$Zn ^{2+}$ is diamagnetic while $Cu ^{2+}$ is paramagnetic, why?

Answer

$Zn ^{+2}\left(3 d^{10}\right)$ does not have a single unpaired electron whereas $Cu ^{2+}\left(3 d^9\right)$ has one unpaired electron present, hence $Zn ^{2+}$ is diamagnetic but $Cu ^{2+}$ is paramagnetic.

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Question 171 Mark

Which of the following oxides will have maximum covalent property? $SC _2 O _3, TiO _2, Mn _2 O _7$ and $V _2 O _5$.

Answer

$Mn _2 O _7$

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Question 181 Mark

Name those elements of the first transition series which show only one type of oxidation state.

Answer

Sc and Zn .

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Question 191 Mark

Write the electronic configuration of $Ni ^{2+}$.

Answer

$Ni ^{2+}=1 s^2 2 s^2 2 p ^6 3 s^2 3 p ^6 3 d^8 4 s^0$

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Question 201 Mark

Actinoid contraction is more than lanthanoid contraction, why?

Answer

Due to weak shielding effect of 5 f electrons, actinoid contraction is more than lanthanoid contraction.

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Question 211 Mark

Instruments used in surgery are cleaned by $KMnO _4$, why?

Answer

Due to the germicidal properties of $KMnO _4$, it can be treated surgically. That is why it is used to clean the equipment.

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Question 221 Mark

Which has the lowest melting point among the elements of $d$-block?

Answer

Mercury (Hg).

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Question 231 Mark

What are parauranium elements?

Answer

Radioactive and artificial elements U are called parauranium elements.

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Question 241 Mark

Write the maximum oxidation state displayed by actinoids.

Answer

Actinoids exhibit maximum oxidation state of $+7$.

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Question 251 Mark

$La ( OH )_3$ is more basic than $Lu ( OH )_3$, why?

Answer

From La to Lu , the alkaline nature of hydroxide decreases because their ionic radius decreases, hence $La ( OH )_3$ is more basic than $Lu ( OH )_3$.

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Question 261 Mark

What is called lanthanoid contraction?

Answer

As the atomic number increases from La to Lu , there is a decrease in the atomic and ionic radii, this is called lanthanoid contraction.

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Question 271 Mark

What are coin metals?

Answer

Copper, Silver and Gold.

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Question 281 Mark

Write equations showing the products formed when dichromate ion is reacted with hyrogen peroxide.

Answer

$Cr _2 O _7^{2-}+4 H _2 O _2+2 \stackrel{+}{ H } \rightarrow 2 CrO _5+5 H _2 O$

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Question 291 Mark

Explain the use of Bayer reagent.

Answer

Unsaturated is tested with Bayer's reagent (1\% alkaline $KMnO _4$ ).

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Question 301 Mark

Write the products formed when KI solution is treated with acidic and basic $KMnO _4$.

Answer

When KI solution is reacted with acidic $KMnO _4$, iodine $\left( I _2\right)$ is obtained and when it is reacted with alkaline $KMnO _4$, potassium iodate $\left( KIO _3\right)$ is obtained.

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Question 311 Mark

Orange solution of $K _2 Cr _2 O _7$ contains strong alkali $( NaOH$ or KOH $)$ the solution turns yellow. Why?

Answer

On adding strong alkali to the orange solution of $K _2 Cr _2 O _7$, chromate is formed, hence the solution turns yellow.
$
\underset{\text { orange }}{Cr_2 O_7^{2-}}+2 \overline{O} H \rightarrow \underset{\text { yellow }}{2 CrO_4^{2-}}+H_2 O
$

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Question 321 Mark

What is the structure of $CrO _4{ }^{2-}$ ion?

Answer

The structure of $CrO _4{ }^{2-}$ is tetrahedral.

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Question 331 Mark

Write the equation of disproportionation of $Mn ^{ VI }$

Answer

$3 \stackrel{ VI }{ MnO _4^{2-}}+4 H ^{+} \rightarrow 2 \stackrel{ VII }{ Mn O _4^{-}}+\stackrel{ VV }{ MnO _2}+2 H _2 O$

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Question 341 Mark