Question

h2o nh4+ brf5 bf3 nh3 co2 ch4 sf4
how many of the molecules above are tetrahedral compounds?
(a number)
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Explanation:

Step1: Analyze each molecule

• \(H_2O\): The central atom \(O\) has \(2\) bonding pairs and \(2\) lone pairs. Using VSEPR theory, the electron - pair geometry is tetrahedral, but the molecular geometry (due to lone pairs) is bent.
• \(NH_4^+\): The central atom \(N\) has \(4\) bonding pairs and \(0\) lone pairs. According to VSEPR theory, the molecular geometry is tetrahedral.
• \(BrF_5\): The central atom \(Br\) has \(5\) bonding pairs and \(1\) lone pair. The electron - pair geometry is octahedral, and the molecular geometry is square - pyramidal.
• \(BF_3\): The central atom \(B\) has \(3\) bonding pairs and \(0\) lone pairs. The molecular geometry is trigonal planar.
• \(NH_3\): The central atom \(N\) has \(3\) bonding pairs and \(1\) lone pair. The electron - pair geometry is tetrahedral, but the molecular geometry (due to lone pair) is trigonal pyramidal.
• \(CO_2\): The central atom \(C\) has \(2\) double - bonding pairs (or \(2\) bonding regions) and \(0\) lone pairs. The molecular geometry is linear.
• \(CH_4\): The central atom \(C\) has \(4\) bonding pairs and \(0\) lone pairs. According to VSEPR theory, the molecular geometry is tetrahedral.
• \(SF_4\): The central atom \(S\) has \(4\) bonding pairs and \(1\) lone pair. The electron - pair geometry is trigonal bipyramidal, and the molecular geometry is see - saw.

Answer:

\(2\) (because \(NH_4^+\) and \(CH_4\) have tetrahedral molecular geometry)