Question:
$PH_4I + NaOH$ forms
$PH_4I + NaOH$ forms
Updated On: Apr 27, 2024
- $PH_3$
- $NH_3$
- $P_4O_6$
- $P_4O_10$
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The Correct Option is A
Approach Solution - 1
When $PH_4I$ and NaOH react, phosphine gas is
obtained.
$PH_4I + NaOH \, \longrightarrow\, PH_3 + Nal + H_2O$
obtained.
$PH_4I + NaOH \, \longrightarrow\, PH_3 + Nal + H_2O$
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Approach Solution -2
To solve this problem, we will first see what is actually happening in this reaction.
Then we look at it with the help of the reaction and finally, we will analyze the final product.
So let's understand that the reactants are phosphonium and sodium hydroxide and by heating these we get phosphine gas, water, and sodium iodide.
The reaction is below :
PH4I+NaOH→PH3+H2O+NaI
So, this is the final reaction of the given half-reaction. So, the correct answer is phosphine gas.
So, the correct option is A.
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Concepts Used:
P-Block Elements
- P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
- P block elements are shiny and usually a good conductor of electricity and heat as they have a tendency to lose an electron. You will find some amazing properties of elements in a P-block element like gallium. It’s a metal that can melt in the palm of your hand. Silicon is also one of the most important metalloids of the p-block group as it is an important component of glass.
P block elements consist of:
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