Classification of Elements and Periodicity in Properties
89320
The valency shell of element $A$ contains 3 electrons while the valency shell of element $B$ contains 6 electrons. If $A$ combines with $B$, the probable formula of the compound formed will be
1 $\mathrm{AB}_2$
2 $\mathrm{A}_2 \mathrm{~B}$
3 $\mathrm{A}_2 \mathrm{~B}_3$
4 $\mathrm{A}_3 \mathrm{~B}_2$
Explanation:
The element $\mathrm{A}$ is $\mathrm{ns}^2 \mathrm{np}^1$ and $B$ is $\mathrm{ns}^2 \mathrm{np}^4$. They can form compound of the type $\mathrm{A}_2 \mathrm{~B}_3$.
BITSAT-2014
Classification of Elements and Periodicity in Properties
89321
Variable valency is a general feature of :
1 s-block elements
2 p-block elements
3 d-block elements
4 (b) and (c) both
Explanation:
Variable valency is the feature of d-block element. Variation in valency is due to participation of $(\mathrm{n}-1)$ d electron in bond formation. d-block elements have empty inner orbitals, the electrons are added to incomplete inner shells.
MPPET-2013
Classification of Elements and Periodicity in Properties
89310
Identify the element for which electronic configuration in +3 oxidation state is $[\mathrm{Ar}] 3 \mathrm{~d}^5$.
1 $\mathrm{Ru}$
2 $\mathrm{Mn}$
3 $\mathrm{Co}$
4 $\mathrm{Fe}$
Explanation:
The atomic number of $\mathrm{Ar}$ is 18 due to presence of electrons in d-orbital's and add 5 electron in the d-orbital. Total number of electrons in $\mathrm{X}^{3+}$ is 23 . Oxidation means loss of electrons, so +3 charges implies 3 electrons loss by $\mathrm{X}$ to form $\mathrm{X}^{3+}$. Hence, the number of electrons in $X=23+3=26$ Therefore the element is $\mathrm{Fe}$.
(JEE Main 2021
Classification of Elements and Periodicity in Properties
89316
The element that forms stable compounds in low oxidation state is
1 $\mathrm{Mg}$
2 $\mathrm{Al}$
3 $\mathrm{Ga}$
4 $\mathrm{Tl}$
Explanation:
The element that forms stable compound in low oxidation state is Thallium (Tl) due to inert pair effect, and poor shielding. Increased effective nuclear charge holds the $6 \mathrm{~s}$ electron tightly.
Classification of Elements and Periodicity in Properties
89320
The valency shell of element $A$ contains 3 electrons while the valency shell of element $B$ contains 6 electrons. If $A$ combines with $B$, the probable formula of the compound formed will be
1 $\mathrm{AB}_2$
2 $\mathrm{A}_2 \mathrm{~B}$
3 $\mathrm{A}_2 \mathrm{~B}_3$
4 $\mathrm{A}_3 \mathrm{~B}_2$
Explanation:
The element $\mathrm{A}$ is $\mathrm{ns}^2 \mathrm{np}^1$ and $B$ is $\mathrm{ns}^2 \mathrm{np}^4$. They can form compound of the type $\mathrm{A}_2 \mathrm{~B}_3$.
BITSAT-2014
Classification of Elements and Periodicity in Properties
89321
Variable valency is a general feature of :
1 s-block elements
2 p-block elements
3 d-block elements
4 (b) and (c) both
Explanation:
Variable valency is the feature of d-block element. Variation in valency is due to participation of $(\mathrm{n}-1)$ d electron in bond formation. d-block elements have empty inner orbitals, the electrons are added to incomplete inner shells.
MPPET-2013
Classification of Elements and Periodicity in Properties
89310
Identify the element for which electronic configuration in +3 oxidation state is $[\mathrm{Ar}] 3 \mathrm{~d}^5$.
1 $\mathrm{Ru}$
2 $\mathrm{Mn}$
3 $\mathrm{Co}$
4 $\mathrm{Fe}$
Explanation:
The atomic number of $\mathrm{Ar}$ is 18 due to presence of electrons in d-orbital's and add 5 electron in the d-orbital. Total number of electrons in $\mathrm{X}^{3+}$ is 23 . Oxidation means loss of electrons, so +3 charges implies 3 electrons loss by $\mathrm{X}$ to form $\mathrm{X}^{3+}$. Hence, the number of electrons in $X=23+3=26$ Therefore the element is $\mathrm{Fe}$.
(JEE Main 2021
Classification of Elements and Periodicity in Properties
89316
The element that forms stable compounds in low oxidation state is
1 $\mathrm{Mg}$
2 $\mathrm{Al}$
3 $\mathrm{Ga}$
4 $\mathrm{Tl}$
Explanation:
The element that forms stable compound in low oxidation state is Thallium (Tl) due to inert pair effect, and poor shielding. Increased effective nuclear charge holds the $6 \mathrm{~s}$ electron tightly.
Classification of Elements and Periodicity in Properties
89320
The valency shell of element $A$ contains 3 electrons while the valency shell of element $B$ contains 6 electrons. If $A$ combines with $B$, the probable formula of the compound formed will be
1 $\mathrm{AB}_2$
2 $\mathrm{A}_2 \mathrm{~B}$
3 $\mathrm{A}_2 \mathrm{~B}_3$
4 $\mathrm{A}_3 \mathrm{~B}_2$
Explanation:
The element $\mathrm{A}$ is $\mathrm{ns}^2 \mathrm{np}^1$ and $B$ is $\mathrm{ns}^2 \mathrm{np}^4$. They can form compound of the type $\mathrm{A}_2 \mathrm{~B}_3$.
BITSAT-2014
Classification of Elements and Periodicity in Properties
89321
Variable valency is a general feature of :
1 s-block elements
2 p-block elements
3 d-block elements
4 (b) and (c) both
Explanation:
Variable valency is the feature of d-block element. Variation in valency is due to participation of $(\mathrm{n}-1)$ d electron in bond formation. d-block elements have empty inner orbitals, the electrons are added to incomplete inner shells.
MPPET-2013
Classification of Elements and Periodicity in Properties
89310
Identify the element for which electronic configuration in +3 oxidation state is $[\mathrm{Ar}] 3 \mathrm{~d}^5$.
1 $\mathrm{Ru}$
2 $\mathrm{Mn}$
3 $\mathrm{Co}$
4 $\mathrm{Fe}$
Explanation:
The atomic number of $\mathrm{Ar}$ is 18 due to presence of electrons in d-orbital's and add 5 electron in the d-orbital. Total number of electrons in $\mathrm{X}^{3+}$ is 23 . Oxidation means loss of electrons, so +3 charges implies 3 electrons loss by $\mathrm{X}$ to form $\mathrm{X}^{3+}$. Hence, the number of electrons in $X=23+3=26$ Therefore the element is $\mathrm{Fe}$.
(JEE Main 2021
Classification of Elements and Periodicity in Properties
89316
The element that forms stable compounds in low oxidation state is
1 $\mathrm{Mg}$
2 $\mathrm{Al}$
3 $\mathrm{Ga}$
4 $\mathrm{Tl}$
Explanation:
The element that forms stable compound in low oxidation state is Thallium (Tl) due to inert pair effect, and poor shielding. Increased effective nuclear charge holds the $6 \mathrm{~s}$ electron tightly.
Classification of Elements and Periodicity in Properties
89320
The valency shell of element $A$ contains 3 electrons while the valency shell of element $B$ contains 6 electrons. If $A$ combines with $B$, the probable formula of the compound formed will be
1 $\mathrm{AB}_2$
2 $\mathrm{A}_2 \mathrm{~B}$
3 $\mathrm{A}_2 \mathrm{~B}_3$
4 $\mathrm{A}_3 \mathrm{~B}_2$
Explanation:
The element $\mathrm{A}$ is $\mathrm{ns}^2 \mathrm{np}^1$ and $B$ is $\mathrm{ns}^2 \mathrm{np}^4$. They can form compound of the type $\mathrm{A}_2 \mathrm{~B}_3$.
BITSAT-2014
Classification of Elements and Periodicity in Properties
89321
Variable valency is a general feature of :
1 s-block elements
2 p-block elements
3 d-block elements
4 (b) and (c) both
Explanation:
Variable valency is the feature of d-block element. Variation in valency is due to participation of $(\mathrm{n}-1)$ d electron in bond formation. d-block elements have empty inner orbitals, the electrons are added to incomplete inner shells.
MPPET-2013
Classification of Elements and Periodicity in Properties
89310
Identify the element for which electronic configuration in +3 oxidation state is $[\mathrm{Ar}] 3 \mathrm{~d}^5$.
1 $\mathrm{Ru}$
2 $\mathrm{Mn}$
3 $\mathrm{Co}$
4 $\mathrm{Fe}$
Explanation:
The atomic number of $\mathrm{Ar}$ is 18 due to presence of electrons in d-orbital's and add 5 electron in the d-orbital. Total number of electrons in $\mathrm{X}^{3+}$ is 23 . Oxidation means loss of electrons, so +3 charges implies 3 electrons loss by $\mathrm{X}$ to form $\mathrm{X}^{3+}$. Hence, the number of electrons in $X=23+3=26$ Therefore the element is $\mathrm{Fe}$.
(JEE Main 2021
Classification of Elements and Periodicity in Properties
89316
The element that forms stable compounds in low oxidation state is
1 $\mathrm{Mg}$
2 $\mathrm{Al}$
3 $\mathrm{Ga}$
4 $\mathrm{Tl}$
Explanation:
The element that forms stable compound in low oxidation state is Thallium (Tl) due to inert pair effect, and poor shielding. Increased effective nuclear charge holds the $6 \mathrm{~s}$ electron tightly.
