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NEET Practice Questions (MCQs) with Answers & Solutions

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Heating of Ag with dil. HNO₃ give

NO₂

N₂O

N₂O₃

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Explanation

$2 H N O_{3} (d i l .) \to 2 N O + H_{2} O + 3 O 2 A g + 2 H N O_{3} + O \to 2 A g N O_{3} + H_{2} O] \times 3 6 A g + 8 H N O_{3} \to 6 A g N O_{3} + 2 N O + 4 H_{2} O$

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Zinc dissolve in an excess of NaOH, because of the formation of

ZnO

Zn(OH)₂

NaZn(OH)₃

Na₂ZnO₂

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Explanation

It react with NaOH on moderate heating & form soluble Zincate.

$Z n + 2 N a O H \to N a_{2} Z n O_{2} + H_{2}$

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When excess of SnCl₂ is added to HgCl₂, the substance formed is

Hg₂Cl₂

Cl₂

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Explanation

$S n C l_{2} r e d u c e s H g C l_{2} t o H g_{2} C l_{2} a n d f i n a l l y t o H g 2 H g C l_{2} + S n C l_{2} \to H g_{2} C l_{2} (w h i t e p p t) + S n C l_{4} H g_{2} C l_{2} + S n C l_{2} \to 2 H g (b l a c k) + S n C l_{4}$

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The oxidation number is changed in which of the following case

SO₂ gas is passed into $C r_{2} O_{7}^{2 -} / H^{+}$

Aqueous solution of $C r O_{4}^{2 -}$ is acidified

$C r O_{2} C l_{2}$ is dissolved in NaOH

$C r_{2} O_{7}^{2 -}$ solution is made alkaline

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Explanation

$(A) C r_{2} O_{7}^{2 -} + S O_{2} + 2 H^{+} \to 2 C r^{3 +} + 3 S O_{4}^{2 -} + H_{2} O (c h a n g e) + 6 + 4 + 3 + 6 (B) 2 C r_{2} O_{4}^{2 -} + 2 H^{+} \to C r_{2} O_{7}^{2 -} + H_{2} O (N o c h a n g e) + 6 + 6 (C) C r O_{2} C l_{2} + 2 N a O H \to N a C r O_{4} (N o c h a n g e) + 6 + 6 (D) C r_{2} O_{7}^{2 -} + 2 O H^{-} \to 2 C r O_{4}^{2 -} + H_{2} O (N o c h a n g e) + 6 + 6$

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Which of the following elements form stable dinuclear ions

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Explanation

Hg(l) compound contain mercurous ions (Hg-Hg)²⁺ & not Hg⁺.

The two atoms are bonded together using 6s-orbital.

Zn & Cd also form dinuclear ion (Zn-Zn)²⁺ & (Cd-Cd)²⁺ ; however these ions are unstable & have only been detected in melts of Zn/ZnCl₂ &Cd/CdCl₂ respectively.

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A mixture of Mn²⁺ & Zn²⁺ can be separated by using an excess of

NH₄OH

NaOH

H₂SO₄

HNO₃

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Explanation

$M n^{2 +} + 2 N a O H \to M n (O H)_{2} + 2 N a^{+} (p p t) Z n^{2 +} + 2 N a O H \to Z n (O H)_{2} + 2 N a^{+} (p p t) Z n (O H)_{2} + 2 N a O H \to N a_{2} [Z n (O H)_{2}] S o l u b l e c o m p l e x$

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The correct electronic configuration of Ti(Z=22 atom) is

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2}$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 p^{6}$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{4}$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{2}$

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Explanation

Ti(Z=22) is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{2}$

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Of the ions, Zn²⁺, Ni²⁺and Cr³⁺(atomic number of Zn =30, Ni=28 and Cr=24).

Only Zn²⁺is Colourless and Ni²⁺and Cr³⁺are coloured

All these are colourless

All these are coloured

Only Ni²⁺is coloured and Zn²⁺and Cr³⁺are colourless.

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Explanation

Zn²⁺is Colourless due to absence of any unpaired electron.Ni²⁺and Cr³⁺are coloured due to 2 and 3 unpaired electrons respecively.

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Which is not true in case of transition metals

They are malleable and ductile

They have high melting and boiling points

They crystallise with body centered cubic and hexagonal close packed structure only

They show variable oxidation states although not always

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Explanation

The transition elements exhibit hexagonal closed packed,face centred cube and body centred cubic structures and co-ordination number range from 8 to 12.

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Cerium(Z=58) is an important member of lanthanoids. Which of the following statements about cerium is incorrect

The common oxidation states of cerium are +3 and +4.

Cerium (IV) acts as an oxidising agent

The +4 oxidation state of cerium is not known in solutions

The +3 oxidation state of cerium is more stable than the +4 oxidation state.

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Explanation

Ce(IV) is a stable species in solution.

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