Q: Which of the following hydrides is electron-precise hydride?

Methane \$CH_4\$ is electron-precise hydride.Diborane \$B_2H_6\$ is electron-deficient hydrideAmmonia \$NH_3\$ and water \$H_2O\$ are electron-rich hydrides.Elements of group 14 form electron-precise (having required number of electrons to write the Lewis structure ) form precise hydrides.

Q: Elements of which of the following groups do not form hydrides?

The region of the periodic table from the group \$7\$ to \$9\$ is referred to as the hydride gap as they do not form hydrides. Examples of such elements are Mn, Fe, Co, Ru etc. These elements do not form hydrides on account of low affinity for hydrogen in their normal oxidation states. Hence, option C is correct.

Q: Which of the following hydrides is electron deficient?

\$ \left (BH_{3} \right )\$ is the chemical formula for the Boron atom. It is surrounded by 6 electrons, so it is electron deficient due to its incomplete octet. B-atom in diborane is \$sp^{3}\$ hybridized. \$B_{2}H_{6}\$ molecule has a total of 8 covalent bonds but only 12 electrons i.e. 6 pairs. So, it has two electron pairs less than the maximum required number for bonding, which makes it electron deficient. The answer is option D.

Q: Among \$CaH_2, NH_3, NaH\$ and \$B_2H_6\$, which are covalent hydrides?

\$NH_3\$ and \$B_2H_6\$ are the covalent hydrides and \$CaH_2\$ and \$NaH\$ are the ionic hydride.\$N\$ and \$B\$ form covalent hydride as they have high electronegativity and small size.

Q: What is hydride? How many types of hydrides are there? Explain example.

Hydrides The binary compounds of dihydrogen with oredr elements except noble gases are called HydridesThere are following types of hydrides:1. Covalent of Molecular Hydride: Dehydrogen forms molecular hydrides with most of the \$p-\$block elements. They consist of discrete covalent molecules with are held together by weak Vander Waal's forces of attraction. They have low melting and boiling point. They are volatile in naturee.g., \$CH_4, NH_3, H_2O\$ and \$HF\$.These are of three types on the basis of relative number of electrons and bonds in their Lewis structure:\$\bullet\$ Electron deficient hydride: Electrons of group \$13\$ will electrons deficient hydride. They act as Lewis acids, e.g. \$B_2H_6\$\$\bullet\$ Electrons-precise hydride: All elements group \$14\$ from electrons presise hydrides e.g, \$CH_4\$\$\bullet\$ Electrons hydride: Elements of group \$15-17\$ from such compounds. They act as Lewis bases e.g \$NH_3\$2. Ionic or Saline hydride: These are stoichiometic compounds of hydrogen formed with most of the \$s-\$block elements. i.e., alkali and alkaline earth metals. These hydride are crystalline, non volatile and non-conducting solid state. They have high melting and boiling points.e.e., \$NaH, BeH_2, MgH_2\$ etc.3. Metallic or Interstitial hydrides: These hydrides are formed by the reaction of dihydrogen with \$d-\$block and \$f-\$block elements. Due to small size dihydrogen in these hydride, hydrogen occupies interstices in the metal lattice producing distribution without any charge in its type. So these are known as interstitial hydrides. These are non-stoichiometric being deficient in hydrogen.e.g., \$scH_2, LaH_2, T_2H_2, CrH_3\$ etc.

Q: Ionic hydride reacts with water to give:

Ionic hydrides give basic solution when reacts with water. e.g.\$LiH+{ H }_{ 2 }O\longrightarrow \underset{base}{LiOH}+{ H }_{ 2 }\uparrow \$

Q: Non-stoichiometric hydrides are produced by:

The hydrogen deficient compounds formed by the reaction of \$d-block\$ and \$f-block\$ elements with dihydrogen are called Non-stoichiometric compounds. The d-block and f-block element form non-stoichiometric hydride because of the vacant d- and f-orbitals along with the small size.Their elemental composition proportions cannot be represented in integers. They disobey the law of constant composition. Among the elements given, only vanadium and palladium form non-stoichiometric hydrides.Answer: (A) palladium, vanadium

Q: Metal hydrides are ionic, covalent or molecular in nature. Among \$LiH,\ NaH,\ KH,\ RbH\$ and \$ CsH\$, the correct order of increasing ionic character is:

According to Fajan’s rule, as the size of cation increases, the polarisation of anion decreases; the percentage covalent character decreases and the percentage ionic character increases.And as we move down the group, the size of elements increases .Thus, the order of atomic radii of the alkali metals is \$Li < Na< K< Rb< Cs\$.Thus, order of increasing percentage ionic character in alkali metal hydrides will be: \$LiH< NaH<KH< RbH<CsH\$

Q: The hydride ion \$H^{-}\$ is stronger base than hydroxide ion \$OH^{-}\$. Which of the following reaction will occur if sodium hydride \$(NaH)\$ is dissolved in water?

As hydride ion is stronger base, so it will react with water which acts as an acid. The reaction occurring is as follows: \$H^{-}(aq.)+H_{2}O(l) \rightarrow OH^{-}+H_{2}\$.

Question 1

Which of the following hydrides is electron-precise hydride?

Accepted Answer

Methane $CH_4$ is electron-precise hydride.Diborane $B_2H_6$ is electron-deficient hydrideAmmonia $NH_3$ and water $H_2O$ are electron-rich hydrides.Elements of group 14 form electron-precise (having required number of electrons to write the Lewis structure ) form precise hydrides.

Question 2

Elements of which of the following groups do not form hydrides?

Accepted Answer

The region of the periodic table from the group $7$ to $9$ is referred to as the hydride gap as they do not form hydrides. Examples of such elements are Mn, Fe, Co, Ru etc. These elements do not form hydrides on account of low affinity for hydrogen in their normal oxidation states. Hence, option C is correct.

Question 3

Arrange the following.(i) $CaH_2$, $BeH_2$ and $TiH_2$ in order of increasing electrical conductance.(ii) $LiH, NaH$ and $CsH$ in order of increasing ionic character.(iii) $H-H, D-D$ and $F-F$ in order of increasing bond dissociation enthalpy.(iv) $NaH, MgH_2$ and $H_2O$ in order of increasing reducing property.

Accepted Answer

(i) Ionic compounds conduct electricity whereas covalent compounds does not.$\displaystyle BeH_2 $ is a covalent hydride and does not conduct electricity. $\displaystyle CaH_2 $ is an ionic hydride and conducts electricity in molten state. $\displaystyle TiH_2 $ is metallic in nature and conducts electricity at room temperature.Hence, the increasing order of electrical conductivity is $\displaystyle BeH_2<CaH_2<TiH_2 $.(ii) The ionic character of bond depends on electronegativity difference between two atoms. When the electronegativity difference is larger, the ionic character is smaller. On moving down the group of alkali metals, the electronegativity decreases from Li to Cs. Hence the ionic character increases in the order $\displaystyle LiH < NaH < CsH $.(iii) Bond dissociation energy depends on bond strength. Bond strength depends on the attractive and repulsion forces present in a molecule.Due to higher nuclear mass of $\displaystyle D_2 $, the attraction between nucleus and bond pair in D-D is stronger than in H-H. This results in greater bond strength and higher bond dissociation enthalpy. Thus, the bond dissociation enthalpy of $\displaystyle D-D $ is higher than that of $\displaystyle H-H $.The bond dissociation enthalpy of F-F is minimum as the repulsion between the bond pair and lone pairs of F is strong. Hence, the increasing order of bond dissociation enthalpy is $\displaystyle F-F < H-H < D-D $.(iv) NaH is an ionic hydride and can easily donate its electrons. It is most reducing. $\displaystyle MgH_2 $ and $\displaystyle H_2O $ are covalent hydrides. $\displaystyle H_2O $ has lower reducing power than $\displaystyle MgH_2 $ as its bond dissociation energy is higher. Hence, the increasing order of the reducing property is $\displaystyle H_2O < MgH_2 < NaH $.

Question 4

Which of the following hydrides is electron deficient?

Accepted Answer

$ \left (BH_{3} \right )$ is the chemical formula for the&#10;Boron atom. It is surrounded by 6 electrons, so it is electron deficient due to&#10;its incomplete octet. B-atom in diborane is $sp^{3}$ hybridized. $B_{2}H_{6}$&#10;molecule has a total of 8 covalent bonds but only 12 electrons i.e. 6 pairs. So, it&#10;has two electron pairs less than the maximum required number for bonding, which&#10;makes it electron deficient.&#10;&#10; The answer is option D.

Question 5

Among $CaH_2, NH_3, NaH$ and $B_2H_6$, which are covalent hydrides?

Accepted Answer

$NH_3$ and $B_2H_6$ are the covalent hydrides and $CaH_2$ and $NaH$ are the ionic hydride.$N$ and $B$ form covalent hydride as they have high electronegativity and small size.

Question 6

What is hydride? How many types of hydrides are there? Explain example.

Accepted Answer

Hydrides The binary compounds of dihydrogen with oredr elements except noble gases are called HydridesThere are following types of hydrides:1. Covalent of Molecular Hydride: Dehydrogen forms molecular hydrides with most of the $p-$block elements. They consist of discrete covalent molecules with are held together by weak Vander Waal's forces of attraction. They have low melting and boiling point. They are volatile in naturee.g., $CH_4, NH_3, H_2O$ and $HF$.These are of three types on the basis of relative number of electrons and bonds in their Lewis structure:$\bullet$ Electron deficient hydride: Electrons of group $13$ will electrons deficient hydride. They act as Lewis acids, e.g. $B_2H_6$$\bullet$ Electrons-precise hydride: All elements group $14$ from electrons presise hydrides e.g, $CH_4$$\bullet$ Electrons hydride: Elements of group $15-17$ from such compounds. They act as Lewis bases e.g $NH_3$2. Ionic or Saline hydride: These are stoichiometic compounds of hydrogen formed with most of the $s-$block elements. i.e., alkali and alkaline earth metals. These hydride are crystalline, non volatile and non-conducting solid state. They have high melting and boiling points.e.e., $NaH, BeH_2, MgH_2$ etc.3. Metallic or Interstitial hydrides: These hydrides are formed by the reaction of dihydrogen with $d-$block and $f-$block elements. Due to small size dihydrogen in these hydride, hydrogen occupies interstices in the metal lattice producing distribution without any charge in its type. So these are known as interstitial hydrides. These are non-stoichiometric being deficient in hydrogen.e.g., $scH_2, LaH_2, T_2H_2, CrH_3$ etc.

Question 7

Ionic hydride reacts with water to give:

Accepted Answer

Ionic hydrides give basic solution when reacts with water. e.g.$LiH+{ H }_{ 2 }O\longrightarrow \underset{base}{LiOH}+{ H }_{ 2 }\uparrow $

Question 8

Non-stoichiometric hydrides are produced by:

Accepted Answer

The hydrogen deficient compounds formed by the reaction of $d-block$ and $f-block$ elements with dihydrogen are called Non-stoichiometric compounds. The d-block and f-block element form non-stoichiometric hydride because of the vacant d- and f-orbitals along with the small size.Their elemental composition proportions cannot be represented in integers. They disobey the law of constant composition. Among the elements given, only vanadium and palladium form non-stoichiometric hydrides.Answer: (A) palladium, vanadium

Question 9

Metal hydrides are ionic, covalent or molecular in nature. Among $LiH,\ NaH,\ KH,\ RbH$ and $ CsH$, the correct order of increasing ionic character is:

Accepted Answer

According to Fajan&#8217;s rule, as the size of cation increases, the polarisation of anion decreases; the percentage covalent character decreases and the percentage ionic character increases.And as we move down the group, the size of elements increases .Thus, the order of  atomic radii of the alkali metals is $Li < Na< K< Rb< Cs$.Thus, order of increasing percentage ionic character in alkali metal hydrides will be:        $LiH< NaH<KH< RbH<CsH$

Question 10

The hydride ion $H^{-}$ is stronger base than hydroxide ion $OH^{-}$. Which of the following reaction will occur if sodium hydride $(NaH)$ is dissolved in water?

Accepted Answer

As hydride ion is stronger base, so it will react with water which acts as an acid. The reaction occurring is as follows: $H^{-}(aq.)+H_{2}O(l) \rightarrow OH^{-}+H_{2}$.

Revise with Concepts

Hydrides - Ionic, Covalent and Metallic

Learn with Videos

Quick Summary With Stories

Preparation and Characteristics of Ionic Hydrides

Uses of Ionic Hydrides

Covalent Hydrides and their Preparation

Properties of Covalent Hydrides

Metallic Hydrides

Which of the following hydrides is electron-precise hydride?

Elements of which of the following groups do not form hydrides?

Which of the following hydrides is electron deficient?

Among $C a H_{2}, N H_{3}, N a H$ and $B_{2} H_{6}$ , which are covalent hydrides?

What is hydride? How many types of hydrides are there? Explain example.

Ionic hydride reacts with water to give:

Non-stoichiometric hydrides are produced by:

Metal hydrides are ionic, covalent or molecular in nature. Among $L i H, N a H, K H, R b H$ and $C s H$ , the correct order of increasing ionic character is:

The hydride ion $H^{-}$ is stronger base than hydroxide ion $O H^{-}$ . Which of the following reaction will occur if sodium hydride $(N a H)$ is dissolved in water?

Revise with Concepts

Hydrides - Ionic, Covalent and Metallic

Learn with Videos

Quick Summary With Stories

Preparation and Characteristics of Ionic Hydrides

Uses of Ionic Hydrides

Covalent Hydrides and their Preparation

Properties of Covalent Hydrides

Metallic Hydrides

Which of the following hydrides is electron-precise hydride?

Elements of which of the following groups do not form hydrides?

Which of the following hydrides is electron deficient?

Among CaH2​,NH3​,NaH and B2​H6​, which are covalent hydrides?

What is hydride? How many types of hydrides are there? Explain example.

Ionic hydride reacts with water to give:

Non-stoichiometric hydrides are produced by:

Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RbH and CsH, the correct order of increasing ionic character is:

The hydride ion H− is stronger base than hydroxide ion OH−. Which of the following reaction will occur if sodium hydride (NaH) is dissolved in water?

Among $C a H_{2}, N H_{3}, N a H$ and $B_{2} H_{6}$ , which are covalent hydrides?

Metal hydrides are ionic, covalent or molecular in nature. Among $L i H, N a H, K H, R b H$ and $C s H$ , the correct order of increasing ionic character is:

The hydride ion $H^{-}$ is stronger base than hydroxide ion $O H^{-}$ . Which of the following reaction will occur if sodium hydride $(N a H)$ is dissolved in water?