Statement-I : Among \(15^{th}\) group hydrides reducing character decreases from
\(NH_3\) to \(BiH_3\).
Statement-II : \(E_2O_3\) and \(E_2O_5\) are always basic. [Where \(E\) is group \(15\) element]
\(NH_3\) to \(BiH_3\).
Statement-II : \(E_2O_3\) and \(E_2O_5\) are always basic. [Where \(E\) is group \(15\) element]
- Both statement-I and Statement-II are correct
- Statement-I is correct and Statement-II is false
- Statement-I is false and Statement-II is correct
- Both Statement-I and Statement-II are false
The Correct Option is A
Solution and Explanation
The Correct Option is (B) : \(\text{Statement-I is correct and Statement-II is false}\)
Top Questions on hydrides
- What is the order of reducing character for \(AsH_3, NH_3, PH_3\) (group \(15\) hydrides)?
- Bond dissociation energy of "E-H" bond of the " $H _2 E$ " hydrides of group $16$ elements (given below), follows order.
A. $O$ B. $S$ C. $Se$ D. $Te$
Choose the correct from the options given below: - Lithium aluminium hydride can be prepared from the reaction of
- Which hydride among the following is less stable?
- The correct statements among (a) to (b) are: (a) saline hydrides produce $H_2$ gas when reacted with ${H2O}$. (b) reaction of $LiAH_4$ with $BF_3$ leads to $B_2H_6$. (c) $PH_3$ and $CH_4$ are electron - rich and electronprecise hydrides, respectively. (d) HF and $CH_4$ are called as molecular hydrides.
Questions Asked in JEE Main exam
- A convex lens of focal length 40 cm forms an image of an extended source of light on a photo-electric cell. A current \( I \) is produced. The lens is replaced by another convex lens having the same diameter but focal length 20 cm. The photoelectric current now is:
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The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is:
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- Let \[ \vec{a} = 2\hat{i} + \alpha \hat{j} + \hat{k}, \quad \vec{b} = -\hat{i} + \hat{k}, \quad \vec{c} = \beta \hat{j} - \hat{k}, \] where \( \alpha \) and \( \beta \) are integers and \( \alpha \beta = -6 \). Let the values of the ordered pair \( (\alpha, \beta) \) for which the area of the parallelogram of diagonals \( \vec{a} + \vec{b} \) and \( \vec{b} + \vec{c} \) is \( \frac{\sqrt{21}}{2} \), be \( (\alpha_1, \beta_1) \) and \( (\alpha_2, \beta_2) \). Then \( \alpha_1^2 + \beta_1^2 - \alpha_2 \beta_2 \) is equal to:
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Concepts Used:
Hydrides
The ones that are anions of a hydrogen atom are hydrides. Hydrogen can act or respond to different elements of the periodic desk. Nucleophilic, decreasing, and fundamental properties may be found in the interior of the chemicals of hydrogen, and hydride. Hydride compounds are devised with all factors of the periodic table, excluding a few noble gases. Hydrides are common compounds of hydrogen but with lesser electronegative elements.
Types of Hydrides:
Depending upon the evolution of a chemical bond and the elements that can act with the hydrogen atoms, there are three types of hydrides:
- Ionic Hydrides
- Covalent Hydrides
- Metallic or Interstitial Hydrides