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3.40   Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is :

(a) F > Cl > O > N

(b) F > O > Cl > N

(c) Cl > F > O > N

(d) O > F > N > Cl

Answer - (b) The oxidizing property of elements increases from left to right across a period because of the presence of vacant d-orbitals in their valence shells. Thus, we get the decreasing order of oxidizing property as . And the oxidizing character of elements decreases down a group. Thus we get . However, the oxidizing character of Oxygen is more than that of Chlorine. So, . Hence the...

3.39   Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :

(a) B > C > Si > N > F

(b) Si > C > B > N > F

(c) F > N > C > B > Si

(d) F > N > C > Si > B

Answer -  (c) F > N > C > B > Si Because non-metallic character of an elements increases from left to right across the period. And here given position of elements: Boron (B)  -   period and  group. Carbon (C)  -   period and  group. Nitrogen (N)  -   period and  group. Fluorine (F)  -   period and  group. And the Silicon (Si) is present in the  period and   group. Hence the correct order of...

3.38   Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :

(a) B > Al > Mg > K

(b) Al > Mg > B > K

(c) Mg > Al > K > B

(d) K > Mg > Al > B

Answer - (d) K > Mg > Al > B Because the metallic character of an element is the tendency of an element to lose electrons easily. So, moving from left to right across the period metallic character of element decreases. Here, K and Mg are s-block elements while B and Al belong to p-block elements. Hence the order is as follows:  K > Mg > Al > B

3.37   Which one of the following statements is incorrect in relation to ionization enthalpy?

(a) Ionization enthalpy increases for each successive electron.

(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.

(c) End of valence electrons is marked by a big jump in ionization enthalpy.

(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Answer - (d) Because electrons in orbitals bearing a lower n value are more attracted to the nucleus than electrons in orbitals bearing a higher n value. Hence the removal of electrons from orbitals bearing a higher n value is easier than the removal of electrons from orbitals having a lower n value.

3.36  The size of isoelectronic species $- F^-$, Ne and $Na^+$ is affected by

(a) nuclear charge (Z )

(b) valence principal quantum number (n)

(c) electron-electron interaction in the outer orbitals

(d) none of the factors because their size is the same.

Answer - (a) Nuclear charge (Z) Because as the nuclear charge increases the size of an isoelectronic species decreases. Here we have the isoelectronic species: , Ne and   has       has      has   Therefore, the order of the increasing size is as follows:

3.35   Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?

(a) Valence principal quantum number (n)

(b) Nuclear charge (Z )

(c) Nuclear mass

(d) Number of core electrons.

Answer - (c) Nuclear Mass. Valence shell electrons are present in the outermost shell of an atom and nuclear mass does not affect the valence shell because the nuclear mass has so small value that it is considered as negligible.

3.34   Which of the following statements related to the modern periodic table is incorrect?

(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.

(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.

(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.

(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up theelectronic configuration
.

Answer - (b) because d-block has a maximum of 10 electrons in all orbitals in a d-subshell and there are 10 columns. As each subshell can have a maximum of 5-orbitals and 10 electrons, therefore, there are 10 vertical columns in a d-block.

3.33  In the modern periodic table, the period indicates the value of :

(a) Atomic number
(b) Atomic mass
(c) Principal quantum number
(d) Azimuthal quantum number.

Answer - (C) Principal quantum number In the modern periodic table, the period indicates the value of, principal quantum number (n) for the outermost shell or the valence shell.

3.32    Predict the formulas of the stable binary compounds that would be formed by the combination of the                             following pairs of elements.

(f)Element 71 and fluorine

The element with the atomic number 71 is Luteteium (Lu) having valency of 3.  Hence, the formula of the compound is  (Lutetium trifluoride).

3.32   Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.

(e) Phosphorus and fluorine

The stable binary compound that would be formed by the combination of Phosphorus and fluorine are: Phosphorus trifluoride  or Phosphorus pentafluoride .

3.32  Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.

(d)  Silicon and oxygen

The stable binary compound that would be formed by the combination of Silicon and oxygen is: Silicon dioxide  .

3.32 Predict the formulas of the stable binary compounds that would be formed by the combination of the                             following pairs of elements.

(c)    Aluminium and iodine

For Aluminium and iodine, the formula of the stable binary compound is:  (Aluminium iodide) .

3.32  Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.

(b)  Magnesium and nitrogen

For Magnesium and nitrogen, the formula of the stable binary compound is:  (Magnesium nitride) .

3.32  Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.

(a)  Lithium and oxygen

For lithium and oxygen, the formula of the stable binary compound is:  (Lithium oxide)

3.31   The first $\Delta _ i H_1$ and the second $\Delta _ i H_2$  ionization enthalpies (in $KJ mol ^{-1}$ )  $\Delta _{eg }H$ electron gain enthalpy (in $KJ mol ^{-1}$) of a few elements are given below:
Elements           $\Delta H_1 \: \: \: \: \: \: \Delta H_2 \: \: \: \: \: \: \: \Delta _{eg }H$
I                         520             7300         –60
II                        419             3051         –48
III                     1681             3374         –328
IV                     1008            1846         –295
V                      2372            5251           +48
VI                     738              1451            –40

(e) the metal which can form a stable binary halide of the formula $MX_2$ (X=halogen).

The metal which can form a stable binary halide of the formula:  (X=halogen) is the element VI which has low first ionization enthalpy  but higher than that of alkali metals.Hence it is an alkaline earth metal and form binary halide.

3.31 The first $\Delta _ i H_1$ and the second $\Delta _ i H_2$  ionization enthalpies (in $KJ mol ^{-1}$ )  $\Delta _{eg }H$ electron gain enthalpy (in $KJ mol ^{-1}$) of a few elements are given below:
Elements           $\Delta H_1 \: \: \: \: \: \: \Delta H_2 \: \: \: \: \: \: \: \Delta _{eg }H$
I                         520             7300         –60
II                        419             3051         –48
III                     1681             3374         –328
IV                     1008            1846         –295
V                      2372            5251           +48
VI                     738              1451            –40

(d)  the least reactive non-metal.

The element IV has a high negative electron gain enthalpy  but not so high that the first ionization enthalpy  and is the least reactive non-metal.

3.31   The first $\Delta _ i H_1$ and the second $\Delta _ i H_2$  ionization enthalpies (in $KJ mol ^{-1}$ )  $\Delta _{eg }H$ electron gain enthalpy (in $KJ mol ^{-1}$) of a few elements are given below:
Elements           $\Delta H_1 \: \: \: \: \: \: \Delta H_2 \: \: \: \: \: \: \: \Delta _{eg }H$
I                         520             7300         –60
II                        419             3051         –48
III                     1681             3374         –328
IV                     1008            1846         –295
V                      2372            5251           +48
VI                     738              1451            –40

(3) the most reactive non-metal.

The element III which has high first ionization enthalpy  and a very high negative electron gain enthalpy  is the most reactive non-metal.

3.31 The first $\Delta _ i H_1$ and the second $\Delta _ i H_2$  ionization enthalpies (in $KJ mol ^{-1}$ )  $\Delta _{eg }H$ electron gain enthalpy (in $KJ mol ^{-1}$) of a few elements are given below:
Elements           $\Delta H_1 \: \: \: \: \: \: \Delta H_2 \: \: \: \: \: \: \: \Delta _{eg }H$
I                         520             7300         –60
II                        419             3051         –48
III                     1681             3374         –328
IV                     1008            1846         –295
V                      2372            5251           +48
VI                     738              1451            –40

(b) the most reactive metal.

The element II which has the least first ionization enthalpy value  and a low negative electron gain enthalpy value  is the most reactive metal.

3.31 The first $\Delta _ i H_1$ and the second $\Delta _ i H_2$  ionization enthalpies (in $KJ mol ^{-1}$ )  $\Delta _{eg }H$ electron gain enthalpy (in $KJ mol ^{-1}$) of a few elements are given below:
Elements           $\Delta H_1 \: \: \: \: \: \: \Delta H_2 \: \: \: \: \: \: \: \Delta _{eg }H$
I                         520             7300         –60
II                        419             3051         –48
III                     1681             3374         –328
IV                     1008            1846         –295
V                      2372            5251           +48
VI                     738              1451            –40

(a)  the least reactive element.

The element which has highest first ionization enthalpy  and positive electron gain enthalpy  is . And also element V shows similar behaviour like inert gases because of positive electron gain enthalpy.

3.30    Assign the position of the element having outer electronic configuration

(iii)     $( n-2 ) f^ 7 ( n-1 ) d^ 1 ns^2 \: \: for\: \: n = 6$ , in the periodic table.

Given the outer electronic configuration of the element:  . For n=6 hence the element belongs to the sixth period and the last electron goes to the f-orbital, the element is from f-block. We have the group number =3. Electronic configuration:  Hence the element is Gadolinium with Z=64.
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