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Which of the following ions does not liberate hydrogen gas on reaction with dilute acids ?
Option: 1 Ti²⁺
Option: 2 V²⁺
Option: 3 Cr²⁺
Option: 4 Mn²⁺

The reactivity of transition metals decreases from Ti^?????2+ to V^??????2+ Ito Cr^?????2+ and finally Mn²⁺. Therefore, the reactivity of Mn²⁺ is very less. Therefore, Mn^?????2+ does not liberate hydrogen gas on reaction with dilute acids.

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Posted by

vishal kumar

The number of P−OH bonds and the oxidation state of phosphorus atom in pyrophosphoric acid (H₄P₂O₇) respectively are :
Option: 1 four and four
Option: 2 five and four
Option: 3 five and five
Option: 4 four and five

The number of P−OH bonds in pyrophosphoric acid (H₄P₂O_??7) is four.

The oxidation state of the phosphorus atom in H_??????4??P₂O₇

⇒ 4+2x−14=0

x = +5

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vishal kumar

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XeF₆ on partial hydrolysis with water produces a compound ‘X’. The same compound ‘X’ is formed when XeF₆ reacts with silica. The compound ‘X’ is :
Option: 1 XeF₂
Option: 2 XeF₄
Option: 3 XeOF₄
Option: 4 XeO₃

Partial hydrolysis of XeF_????6 gives XeOF₄(compound X)
XeF_????6+H₂O→XeOF₄+2HF

XeF_????6 reacts with silica SiO_????2 to form XeOF_???4 (compound X)
2XeF₆+SiO₂→2XeOF₄+SiF_????4

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vishal kumar

The correct order of catenation is:

Option: 1 C > Sn > Si $\approx$ Ge
Option: 2 C > Si > Ge $\approx$ Sn
Option: 3 Si > Sn > C > Ge
Option: 4 Ge > Sn > Si > C

Catenation/ Self Linkage -

Property of elements to form long chains or rings by self linking of their own atoms through covalent bonds. In the carbon family, it decreases down the group. Only carbon atom form double or triple bong involving $p^{\pi } -p^{\pi }$ multiple bond with itself.

The homo atomic bond energies are as follows :

(i) C-C = 83 kcal / mol

(ii) Si - Si = 54 kcal / mol

(iii) Ge - Ge = 40 kcal / mol

(iv) Sn - Sn = 37 kcal / mol

* Very large difference exists between the bond energies of (C-C) & (Si-Si)

but negligible difference is there for (Ge - Ge ) & (Sn-Sn).

Thus, the correct order is:

C > Si > Ge $\approx$ Sn

Therefore, option (2) is correct.

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Posted by

Ritika Jonwal

JEE Main high-scoring chapters and topics

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Chlorine reacts with hot and concentrated NaOH and produces compounds (X) and (Y). Compound (X) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Y) is:
Option: 1 1.66
Option: 2 2.66
Option: 33.66
Option: 4 4.66

As we have learnt,

Chlorine disproportionates in the presence of hot and concentrated NaOH to form chloride ions and chlorate ions

$\mathrm{Cl_{2}\: +\: 6NaOH(hot\: and\: conc.)\: \rightarrow \: NaClO_{3}\: +\: 5NaCl\: +\: 3H_{2}O}$
Y X

$\mathrm{NaCl\: +\: AgNO_{3}\: \rightarrow \: AgCl(s)\: +\: NaNO_{3}(aq)}$
white(ppt)

The structure of the Chlorate ion $(ClO_3^-)$ is given below:

The average bond order is given as:

$B.O = \frac{5}{3} = 1.66$

Hence, the option number (1) is correct.

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Posted by

Kuldeep Maurya

The atomic radius of Ag is closest to:
Option: 1 Au
Option: 2 Ni
Option: 3 Hg
Option: 4 Cu

Atomic Size/Radii -

Because of the lanthanide contraction, the size of the elements of the 4d series is similar to their corresponding elements of 5d series.

Thus, the atomic radius of Ag is closest to Au.

Therefore, Option(1) is correct.

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Posted by

Kuldeep Maurya

NEET 2024 Most scoring concepts

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A metal (A) on heating in nitrogen gas gives compound B. B on treatment with $H_{2}O$ gives a colourless gas which when passed through $CuSO_{4}$ solution gives a dark blue-violet coloured solution. A and B respectively, are :
Option: 1 $Na$ and $Na_{3}N$
Option: 2 $Mg$ and $Mg_{3}N_{2}$
Option: 3 $Mg$ and $Mg(NO_{3})_{2}$
Option: 4 $Na$ and $NaNO_{3}$

Magnesium reacts with air to form its nitride. The nitride upon hydrolysis yields ammonia which gives a deep blue color solution with $CuSO_4$

$3Mg+N_{2}\rightarrow \underset{(B)}{Mg_{3}N_{2}}\overset{H_{2}O}{\rightarrow}Mg(OH)_{2}+NH_{3}$

$CuSO_4 + 2NH_3 \longrightarrow [Cu(NH_3)_2]SO_4$

Therefore, Option(2) is correct.

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Posted by

vishal kumar

White phosphorous on reaction with concentrated $NaOH$ solution in an inert atmosphere of $CO_{2}$ gives phosphine and compound $(X).$ $(X)$ on acidification with $HCl$ gives compound $(Y).$ The basicity of compound $(Y)$ is :
Option: 1 $4$
Option: 2 $3$
Option: 3 $2$
Option: 4 $1$

Thus, basicity is 1.

Therefore, Option(4) is correct.

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vishal kumar

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Consider the following reactions: $NaCl+K_{2}Cr_{2}O_{7}+H_{2}SO_{4} (conc.)\rightarrow A + side\ products$ $A+NaOH\rightarrow B + side\ products$ $B+ H_{2}SO_{4} (dil.)+H_{2}O_{2}\rightarrow C+ side\ products$ The sum of the total number of atoms in one molecule each of A, B and C is:
Option: 1 18
Option: 216
Option: 3 14
Option: 4 20

(A) = CrO₂Cl₂

(B) = Na₂CrO₄

Total atoms = 5 (A) + 7 (B) + 6 (C) = 18 atoms

Thus, the correct answer is 18.

Therefore, option(1) is correct

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Posted by

Kuldeep Maurya

In the following reactions, products (A) and (B) respectively are: $NaOH + Cl_{2}\rightarrow A +NaCl+ side\ product$ (hot and conc.) $Ca(OH)_{2}+Cl_{2}\rightarrow B + CaCl_2+ side\ product$ (dry)
Option: 1 $NaClO_{3}\ and\ Ca(ClO_{3})_{2}$
Option: 2 $NaOCl\ and\ Ca(OCl)_{2}$
Option: 3 $NaClO_{_{3}}\ and\ Ca(OCl)_{2}$
Option: 4 $NaOCl\ and\ Ca(ClO_{3})_{2}$

As we have learnt,

Chlorine reacts with $NaOH$ and $Ca(OH)_2$ and undergoes disproportionation. The products depend upon the concentration of the base and also the temperature.

The reactions are given as

$2Na(OH) + 3 Cl_2 \longrightarrow NaClO_3 +5 NaCl + 3 H_2O$

(Hot and Conc.)

$2 Ca(OH)_2 + 2 Cl_2 \longrightarrow Ca(OCl)_2 + CaCl_2 + 2 H_2O$

(Dry)

Therefore, Option(4) is correct.

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Posted by

Kuldeep Maurya

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Get Answers to all your Questions

All Questions

Which of the following ions does not liberate hydrogen gas on reaction with dilute acids ? Option: 1 Ti2+ Option: 2 V2+ Option: 3 Cr2+ Option: 4 Mn2+

Posted by

vishal kumar

The number of P−OH bonds and the oxidation state of phosphorus atom in pyrophosphoric acid (H4P2O7) respectively are : Option: 1 four and four Option: 2 five and four Option: 3 five and five Option: 4 four and five

Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

XeF6 on partial hydrolysis with water produces a compound ‘X’. The same compound ‘X’ is formed when XeF6 reacts with silica. The compound ‘X’ is : Option: 1 XeF2 Option: 2 XeF4 Option: 3 XeOF4 Option: 4 XeO3

Posted by

vishal kumar

The correct order of catenation is: Option: 1 C > Sn > Si Ge Option: 2 C > Si > Ge Sn Option: 3 Si > Sn > C > Ge Option: 4 Ge > Sn > Si > C

Posted by

Ritika Jonwal

JEE Main high-scoring chapters and topics

Chlorine reacts with hot and concentrated NaOH and produces compounds (X) and (Y). Compound (X) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Y) is: Option: 1 1.66 Option: 2 2.66 Option: 33.66 Option: 4 4.66

Which of the following ions does not liberate hydrogen gas on reaction with dilute acids ?
Option: 1 Ti²⁺
Option: 2 V²⁺
Option: 3 Cr²⁺
Option: 4 Mn²⁺

The number of P−OH bonds and the oxidation state of phosphorus atom in pyrophosphoric acid (H₄P₂O₇) respectively are :
Option: 1 four and four
Option: 2 five and four
Option: 3 five and five
Option: 4 four and five

XeF₆ on partial hydrolysis with water produces a compound ‘X’. The same compound ‘X’ is formed when XeF₆ reacts with silica. The compound ‘X’ is :
Option: 1 XeF₂
Option: 2 XeF₄
Option: 3 XeOF₄
Option: 4 XeO₃

The correct order of catenation is:

Option: 1 C > Sn > Si $\approx$ Ge
Option: 2 C > Si > Ge $\approx$ Sn
Option: 3 Si > Sn > C > Ge
Option: 4 Ge > Sn > Si > C

Chlorine reacts with hot and concentrated NaOH and produces compounds (X) and (Y). Compound (X) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Y) is:
Option: 1 1.66
Option: 2 2.66
Option: 33.66
Option: 4 4.66

The atomic radius of Ag is closest to:
Option: 1 Au
Option: 2 Ni
Option: 3 Hg
Option: 4 Cu

White phosphorous on reaction with concentrated $NaOH$ solution in an inert atmosphere of $CO_{2}$ gives phosphine and compound $(X).$ $(X)$ on acidification with $HCl$ gives compound $(Y).$ The basicity of compound $(Y)$ is :
Option: 1 $4$
Option: 2 $3$
Option: 3 $2$
Option: 4 $1$