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 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

 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

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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

The increasing order of the atomic radii of the following elements is : (a)     C          (b)     O          (c)     F             (d)     Cl             (e)     Br
Option: 1 (a) < (b)< (c)< (d)<(e)
Option: 2 (c)< (b)< (a)< (d)< (e)
Option: 3 (d)< (c)< (b)< (a)< (e)
Option: 4 (b)< (c)< (d)< (a)< (e)
 

The correct increasing order of the atomic radii of elements is given below:

F< O< C< Cl< Br

Atomic radii increase down a group and it decreases from left to right in a period.

Therefore, Option(2) is correct.

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Posted by

Kuldeep Maurya

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The set that contains the atomic numbers of only transition elements.
 
Option: 1 21,25,42,72
Option: 2 21,32,53,64
Option: 3 9,17,34,38  
Option: 4 37, 42,50,64

The set containing transition elements is:

21, 25, 42, 72

Therefore, Option(1) is correct.

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Posted by

Kuldeep Maurya

Within each pair elements F and Cl, S and Se and Li and Na, respectively, the elements that release more energy upon an electron gain are
Option: 1 Cl, Se and Na
Option: 2 Cl, S and Li
Option: 3 F, S and Li
Option: 4 F, Se and Na
 

From the given pairs, the more energy on gaining of an electron will be released by the following elements:
Cl, S, and Li

Therefore, Option(2) is correct.

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Posted by

Ritika Jonwal

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The atomic number of Unnilunium is ________.
 

Let's break the IUPAC Name-

Unnilunium = Un + nil + Un + ium

Un = 1

nil = 0

un = 1

So, The atomic number will be 101.

Ans = 101

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Posted by

Kuldeep Maurya

The correct order of ionic radii of O^{2-},N^{3-},F^-,Mg^{2+},Na^+,Al^{3+}
Option: 1 N^{3-}<F^-<O^{2-}<Mg^{2+}<Na^+<Al^{3+}
Option: 2 Al^{3+}<Na^+<Mg^{2+}<O^{2-}<F^-<N^{3-}
Option: 3 N^{3-}<O^{2-}<F^-<Na^+<Mg^{2+}<Al^{3+}
Option: 4 Al^{3+}<Mg^{2+}<Na^+<F^-<O^{2-}<N^{3-}

The charge to mass ratio(z/e) of the given species is given as follows:

O2- = z/e = 8/10 = 0.8

N3- = 7/10 = 0.7

F- = 9/10 = 0.9

Mg2+ = 12/10 = 1.2

Na+ = 11/10 = 1.2

Al3+ = 13/10 = 1.3

Now, higher the z/e, smaller is the atomic radii.

Thus, the atomic radii follows the order given below:

Al3+ < Mg2+ < Na+ < F- < O2- < N3-

Therefore, Optioin(4) is correct.

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Posted by

Kuldeep Maurya

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The ionic radii of \mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+} \text {and } \mathrm{Mg}^{2+} are in order:
Option: 1 \mathrm{F}^{-}>\mathrm{O}^{2-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}
Option: 2 \mathrm{O^{2-}>F^{-}>Na^{+}>Mg^{2+}}
Option: 3 \mathrm{Mg^{2+}>Na^{+}>F^{-}>O^{2-}}
Option: 4 \mathrm{O^{2-}>F^{-}>Mg^{2+}>Na^{+}}

We know these about ions-

We know as (z/e) ratio increases size decreases.

Thus correct ionic radii order is

\mathrm{O}^{-2}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}

Therefore, the correct option is (2).

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Posted by

Kuldeep Maurya

The elements with atomic number 101 and 104 belongs to, respectively:
Option: 1 Group 11 and Group 4
Option: 2 Actinoids and Group 6
Option: 3 Actinoids and Group 4
Option: 4 Group 4  and Actinoids

\mathrm{Z}=101 \rightarrow\left[\mathrm{R}_{\mathrm{n}}\right]^{86} 7 \mathrm{s}^{2} 5 \mathrm{f}^{13}\textup{ ----- It is Actinoids}

\mathrm{Z}=104 \rightarrow\left[\mathrm{R}_{\mathrm{n}}\right]^{86} 7 \mathrm{~s}^{2} 5 \mathrm{f}^{14} 6 \mathrm{d}^{2}\textup{ ------ It is 4th group element}

So, Actinoids and Group 4, are correct respectively.

Therefore, the correct option is (3).

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Posted by

Kuldeep Maurya

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