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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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Posted by

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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Posted by

vishal kumar

 Which one of the following is an oxide ?
Option: 1  KO2
Option: 2  BaO2
Option: 3  SiO2
Option: 4 CsO2  
 

SiO2  - oxide

KO???2? , CsO????2  -  superoxides

BaO????2 -  peroxide

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Posted by

vishal kumar

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 The electronic configuration with the highest ionization enthalpy is :
Option: 1  [Ne] 3s2 3p1
Option: 2  [Ne] 3s2 3p2
Option: 3  [Ne] 3s2 3p3
Option: 4 [Ar] 3d10 4s2 4p3
 

The electronic configuration with the highest ionization enthalpy is [Ne]3s23p3. On moving down the group, the  ionization enthalpy decreases. In a period, on moving from left tor fight, the  ionization enthalpy increases. 

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Posted by

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

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The correct order of the atomic radii of C, Cs, Al and S is :
Option: 1 S<C<Cs<Al
Option: 2 C<S<Al<Cs
Option: 3 C<S<Cs<Al
Option: 4 S<C<Al<Cs
 

 

Periodicity of atomic radius and ionic radius in period -

In a period from left to right the effective nuclear charge increases because the next electron fills in the same shell. So the atomic size decrease.

- wherein

Li>Be>B>C>N>O>F

 

 

 

Electronegativity and atomic radius -

The attraction between the outer electrons and the nucleus increases as the atomic radius decreases in a period.

- wherein

Electronegativity\propto\frac{1}{atomic\:radius}

 

 

 

Size of atom and ion in a group -

In a group moving from top to the bottom the number of shell increases.So the atomic size increases.

- wherein

Li<Na<K<Rb<Cs

As we know that

From Left to right in a period size decreases and when going down the group size increases

C< S< Al< Cs

Therefore, Option(2) is correct

  

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Posted by

Ritika Jonwal

The IUPAC symbol for the element with atomic number 119 would be:
Option: 1 uue
Option: 2une
Option: 3 unh
Option: 4 uun

 

Nomenclature of elements with atomic number >100 -

The name is derived directly from the atomic number of the element using the following numerical roots:

0 = nil

1 = un

2 = bi

3 = tri

4 = quad

5 = pent

6 = hex

7 = sept

8 = oct

9 = enn

Eg:

 

Atomic number

Name

Symbol

101

Mendelevium (Unnilunium)

Md (Unu)

102

Nobelium (Unnilbium)

No (Unb)



 

-

uue

1  1  9

Un Un ennium

Therefore, Option(1) is correct.

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Posted by

Ritika Jonwal

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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 electron gain enthalpy(in kJ/mol) of fluorine, chlorine, bromine and iodine respectively are:
Option: 1 -333,\: \: -349,\: \: -325\: \: and\: -296
Option: 2 -296,\: \: -325\: \:-333,\: \:and\: -349
Option: 3 -333,\: \: -325\: \:-349\: \:and\: -296
Option: 4 -349,\: \: -333,\: \: -325\: \: and\: -296
 

Electron Gain Enthalpy or Electron Affinity -

Electron Affinity

Electron gain enthalpy is the energy change that occurs when an electron is added to a neutral gaseous atom to form a negative ion. It is also known as electron affinity.

A(g) + e →A-(g) + \DeltaegH

 

Variation of Electron Affinity

  • The electron gain enthalpy becomes less negative in going from top to bottom in a group.

  • In moving from top to bottom in a group, both the atomic size and the nuclear charge increase. But the effect of the increase in atomic size is more prevalent than the nuclear charge.

  • Halogens have the most negative electron gain enthalpies. In moving down from chlorine to iodine, the electron gain enthalpies become less negative due to the increase in their atomic radii.

  • Chlorine has the most negative electron gain enthalpy value than fluorine. Because fluorine is very small in size due to which there is a very strong inter-electronic repulsion for the incoming electron, thus its electron gain enthalpy is less than chlorine.

 

The electron gain enthalpy values are given below:

Fluorine = -333kJ/mol
Chlorine = -349kJ/mol
Bromine = -325kJ/mol
Iodine = -296kJ/mlol

Therefore, Option(1) is correct.

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Posted by

Kuldeep Maurya

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