Tag: atomic structure and chemical bonding

Questions Related to atomic structure and chemical bonding

Pertaining to metallurgy, when "M" is a metal, its compound $MY _2$ is :

ionic bond atomic structure and chemical bonding chemistry

  1. Covalent

  2. Ionic

  3. Inorganic

  4. None of these

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Correct Option: B
Explanation:

Reaction:

$M-2e^- \rightarrow M^{2+}$

$2Y+2e^- \rightarrow 2Y^-$

$M^{2+} + 2Y^- \rightarrow MY _2$

So, this is an ionic compound.

In electrovalency:

ionic bond atomic structure and chemical bonding chemistry

  1. molecules lose electrons forming atoms

  2. molecules gain electrons forming atoms

  3. atoms share electrons

  4. atoms lose or gain electrons forming ions

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Correct Option: D
Explanation:

Electrovalency is shown by atoms when it looses or gains electrons in order to attain stability. Electrovalency is seen in ionic bonds. Electrovalency is equal to the number of electrons lost or gained by atom to form ion.

Example:- $CaCl _2$
Formation of $Ca^{2 _+}(Cl^-) _2$
Electrovalency of Calcium is $+2$ and is called positive electrovalency.
Electrovalency of Chlorine is $-1$ and is called negative electrovalency.

According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by

ionic bond atomic structure and chemical bonding chemistry

  1. $F = \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{d^2}$

  2. $F = \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+ + r^-)^2}$

  3. $F = {4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{d}$

  4. $F = {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+ + r^-)}$

Show answer
Correct Option: A,B
Explanation:

According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by
$F = \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{d^2}$ or $F = \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+  + r^-)^2}$

Which of the following factor(s) favour the formation of ionic bond?

ionic bond atomic structure and chemical bonding chemistry

  1. The electrostatic attraction between charged ions in the crystal (i.e, lattice energy) should be low.

  2. Both atoms should be equal in size.

  3. The combining elements should differ by at least 2.0 in electronegativity.

  4. The cation and anion should have inert gas electronic configuration.

Show answer
Correct Option: C,D
Explanation:

The following factors favour the formation of ionic bond.
The electrostatic attraction between charged ions in the crystal (i.e, lattice energy) should be high.
One of the atoms (metal) should be large in size. Other atom (non metal) should be small in size.
The combining elements should differ by at least 2.0 in electronegativity.
The cation and anion should have inert gas electronic configuration.

Which of the following represents true statement(s)?

ionic bond atomic structure and chemical bonding chemistry

  1. CsCl is more stable than NaCl because $(r _{Na+} + r _{Cl^-}) < (r _{Cs+} + r _{Cl^-}) $

  2. NaCl is more stable than CsCl because $(r _{Na+} + r _{Cl^-}) < (r _{Cs+} + r _{Cl^-}) $

  3. NaCl is more stable than MgO because the product of $q _1$ and $q _2$ in MgO is nearly four times to that in NaCl.

  4. MgO is more stable than NaCl because the product of $q _1$ and $q _2$ in MgO is nearly four times to that in NaCl.

Show answer
Correct Option: B,D
Explanation:

The options (B) and (D)  represents true statements.
NaCl is more stable than CsCl because $(r _{Na+} + r _{Cl^-}) < (r _{Cs+} + r _{Cl^-}) $
MgO is more stable than NaCl because the product of $q _1$ and $q _2$ in MgO is nearly four times to that in NaCl.
Note:
According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by $F

= \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{d^2}$ or $F =

\dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+  + r^-)^2}$

In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed?

ionic bond atomic structure and chemical bonding chemistry

  1. 1

  2. 2

  3. 3

  4. 4

Show answer
Correct Option: B
Explanation:

Electronic configuration of noble gas is $\mathrm{(2,8)}$. It has $\mathrm{8}$ electron in its outermost orbitals.
Element X has $\mathrm{6}$ electrons in its outermost shell. So, it requires $\mathrm{2}$ more electrons to attain the noble gas configuration.
Hence, option $\mathrm{B}$ is the correct answer.

The charged particles that attract each other to form electrovalent compounds are _________.

ionic bond atomic structure and chemical bonding chemistry

  1. Protons and neutrons

  2. Cation and anion

  3. Electrons

  4. none of these

Show answer
Correct Option: B
Explanation:

Cation$\rightarrow $positively charged

Anion$\rightarrow $negatively charged
           Negative charges attract each other
Cation and onion are the charged particles that attract each other to form electrovalent compounds.