Tag: chemistry

Questions Related to chemistry

The bond order in $N _{2}$ is 

chemistry chemical bonding and structure formal charge basics of chemical bonding types of bonds

  1. 2

  2. 2.5

  3. 1.5

  4. 3

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Correct Option: D

Formal charge on nitrogen in $NO _{3}^{-}$ ion is

chemistry chemical bonding and structure formal charge basics of chemical bonding types of bonds

  1. $+1$

  2. $+2$

  3. $+5$

  4. $Zero$

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Correct Option: A

The formal charge on the O atoms in the ion $[:\ddot{O} = N = \ddot{O}:]^+$ is:

chemistry chemical bonding and structure formal charge basics of chemical bonding types of bonds

  1. $-2$

  2. $-1$

  3. $0$

  4. $+1$

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Correct Option: C
Explanation:

Formal charge$=$Valence electrons-non bonding electrons+no of bonds$ =6-(4+2)=0$

In the cyanide ion the formal negative charge is on:

chemistry chemical bonding and structure formal charge basics of chemical bonding types of bonds

  1. $C$

  2. $N$

  3. both $C$ and $N$

  4. resonate between $C$ and $N$

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Correct Option: D
Explanation:

In the cyanide ion the formal negative charge resonates between $C$ and $N$. Cyanide ion has resonating structure, $-\overset { \ _ }{ C } \equiv N\longleftrightarrow \overset { \ _ }{ N } \underrightarrow { = } C$

In which of the following molecules, is the covalent bond most polar?

chemistry chemical bonding and structure formal charge basics of chemical bonding types of bonds

  1. HI

  2. HBr

  3. HCl

  4. HF

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Correct Option: A
Explanation:

Polar covalent bond shows minimum electronegativity difference between the atoms.

$H=2.1,F=4,Cl=3,Br=2.8,I=2.5.$ 

So, $HI=0.4$ with the minimum difference is the most polar covalent bond.

Hence, option A is correct.

A molecule with highest bond energy is:

chemistry chemical bonding and structure formal charge basics of chemical bonding types of bonds

  1. Fluorine

  2. Chlorine

  3. Bromine

  4. Iodine

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Correct Option: B
Explanation:

Bond dissociation energy in $KJ mol^-1$
$F _2-158$
$Cl _2-244$ 
$Br _2-93$ 
$I _2-151$ 
Hence, in order of decreasing bond strength: 
$Cl _2>Br _2>F _2>I _2$
Fluorine is an anomaly. 

Bond strength decreases from chlorine to iodine as down the group, the atomic size becomes larger and thus the valence electron orbitals become more diffused, causing the overlap of orbitals to become less effective. Therefore the halogen-halogen bond becomes weaker from chlorine to iodine.Fluorine is an exception due to its extremely small size. The F-F bond length is so short that the lone pairs of electrons on the fluorine atoms repel each other and weakens the F-F bond. 

In which of the following molecules partial charge separation does not take place?

chemistry chemical bonding and structure formal charge basics of chemical bonding types of bonds

  1. Chlorine

  2. Hydrochloric acid

  3. Water

  4. Ammonia

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Correct Option: A
Explanation:

Cl molecules are covalent in nature so it is non polar molecule. Non polar compound consists of molecules linked through chemical bonds arranged in such a way that the distribution of charges are symmetrical. Non polar bonds do not exhibit polarity and do not form ions in their solution. These compounds do not dissolve in water which is a polar molecule.

The partial charge separation between the bonded atoms gives rise to ___________ in that compound.

chemistry chemical bonding and structure formal charge basics of chemical bonding types of bonds

  1. covalent character

  2. coordinate character

  3. ionic character

  4. none of these

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Correct Option: C
Explanation:

Partial charge separation between the bonded atoms gives rise to ionic character in the molecule due to the electronegativity difference, a quite small degree of polarization occurs.

What is the formal charge on the chlorine atom in the oxyacid $HOClO _2$ if it contains single bonds ?

chemistry chemical bonding and structure formal charge basics of chemical bonding types of bonds

  1. $-2$

  2. $-1$

  3. $+1$

  4. $+2$

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Correct Option: D
Explanation:

formal charge= valence electrons-(lone electrons+bonds)
$=7-(2+3)=+2$

The life span of a Daniel cell may increased by:

chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

  1. large Cu electrode

  2. lowering of CuSO$ _{4}$ concentration

  3. lowering of ZnSO$ _{4}$ concentration

  4. large zinc electrode

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Correct Option: D
Explanation:

In a daniel cell at anode Zn gets converted to $Zn^{2+}$
$\therefore$ If the size of electrode is increases more zinc can be oxidized,
hence the life span of the cell may be increased

Hence, option d is correct.