Answer
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Hint: For calculation of geometry we much know the hybridization of the central atom. To calculate hybridization we need to know the numbers of lone pair and bond pair electrons are present in the given molecule. Depending on them, the hybridization and hence, geometry will vary.
Complete step by step solution:
To have a trigonal pyramidal geometry we need to have 3 bond pair of electron and 1 lone pair of electron. Trigonal pyramidal geometry is actually formed out of tetrahedral geometry. In tetrahedral geometry all 4 are bond pairs, replacing one bond pair with a lone pair will us trigonal pyramidal geometry. Let us check the bond pair and lone pair in each case:
In \[{\text{B}}{{\text{H}}_3}\] and \[{\text{AlC}}{{\text{l}}_3}\]since 3 hydrogen atoms are attached that means 3 bond pair of electrons are present, but boron and aluminium have no lone pairs. They are group 13 elements and hence have valency 3 and no other electrons. So this option is eliminated.
In \[{{\text{H}}_2}{\text{O}}\] oxygen have 6 electrons, out of which 2 electrons bonded with hydrogen will form 2 bond pair and rest will make 2 lone pair of electrons, which is not required and hence this option stands wrong.
In \[{\text{C}}{{\text{H}}_4}\] carbon has 4 electron and all are bonding so no lone pair are present here.
In \[{\text{N}}{{\text{H}}_3}\] nitrogen has 5 electrons in its valance shell and hence is bonded with 3 hydrogen atoms. That makes 3 bond pair and one lone pair and hence \[{\text{N}}{{\text{H}}_3}\] had trigonal geometry.
Hence correct option is D.
Note: Often geometry and shape are confused. We generally use them as same. Actually geometry of \[{{\text{H}}_2}{\text{O}}\], \[{\text{C}}{{\text{H}}_4}\] and \[{\text{N}}{{\text{H}}_3}\]is same that is tetrahedral but their shapes vary. Water has bent shape where as methane has tetrahedral shape and ammonia has trigonal pyramidal shape. Geometry includes both lone pair and bond pair while shape only includes bond pairs.
Complete step by step solution:
To have a trigonal pyramidal geometry we need to have 3 bond pair of electron and 1 lone pair of electron. Trigonal pyramidal geometry is actually formed out of tetrahedral geometry. In tetrahedral geometry all 4 are bond pairs, replacing one bond pair with a lone pair will us trigonal pyramidal geometry. Let us check the bond pair and lone pair in each case:
In \[{\text{B}}{{\text{H}}_3}\] and \[{\text{AlC}}{{\text{l}}_3}\]since 3 hydrogen atoms are attached that means 3 bond pair of electrons are present, but boron and aluminium have no lone pairs. They are group 13 elements and hence have valency 3 and no other electrons. So this option is eliminated.
In \[{{\text{H}}_2}{\text{O}}\] oxygen have 6 electrons, out of which 2 electrons bonded with hydrogen will form 2 bond pair and rest will make 2 lone pair of electrons, which is not required and hence this option stands wrong.
In \[{\text{C}}{{\text{H}}_4}\] carbon has 4 electron and all are bonding so no lone pair are present here.
In \[{\text{N}}{{\text{H}}_3}\] nitrogen has 5 electrons in its valance shell and hence is bonded with 3 hydrogen atoms. That makes 3 bond pair and one lone pair and hence \[{\text{N}}{{\text{H}}_3}\] had trigonal geometry.
Hence correct option is D.
Note: Often geometry and shape are confused. We generally use them as same. Actually geometry of \[{{\text{H}}_2}{\text{O}}\], \[{\text{C}}{{\text{H}}_4}\] and \[{\text{N}}{{\text{H}}_3}\]is same that is tetrahedral but their shapes vary. Water has bent shape where as methane has tetrahedral shape and ammonia has trigonal pyramidal shape. Geometry includes both lone pair and bond pair while shape only includes bond pairs.
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