Tag: ionic bond

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Questions Related to ionic bond

Multiple choice ionic bond atomic structure and chemical bonding chemistry

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

  1. Covalent

  2. Ionic

  3. Inorganic

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Reaction:

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

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

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

So, this is an ionic compound.

Multiple choice ionic bond atomic structure and chemical bonding chemistry

In electrovalency:

  1. molecules lose electrons forming atoms

  2. molecules gain electrons forming atoms

  3. atoms share electrons

  4. atoms lose or gain electrons forming ions
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice ionic bond atomic structure and chemical bonding chemistry

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

  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. $F = {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+ + r^-)}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
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 $F = \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+  + r^-)^2}$

Multiple choice ionic bond atomic structure and chemical bonding chemistry

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

  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.
Reveal answer Fill a bubble to check yourself
C,D Correct answer
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.

Multiple choice ionic bond atomic structure and chemical bonding chemistry

Which of the following represents true statement(s)?

  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.
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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}$

Multiple choice ionic bond atomic structure and chemical bonding chemistry

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?

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice ionic bond atomic structure and chemical bonding chemistry

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

  1. Protons and neutrons

  2. Cation and anion

  3. Electrons

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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.