Question

drag the tiles to the correct locations. each tile can be used more than once, but not all tiles will be used. one or more locations will remain empty.
nitrosyl fluoride has the chemical formula nof. nitrogen has five valence electrons, oxygen has six, and fluorine has seven. complete the lewis
structure for this covalent compound.

Explanation:

Step1: Determine the valence electrons

• Nitrogen (\(N\)) has \(5\) valence electrons.
• Oxygen (\(O\)) has \(6\) valence electrons.
• Fluorine (\(F\)) has \(7\) valence electrons.
• Total valence electrons for \(NOF\): \(5 + 6+7=18\)

Step2: Arrange the atoms

• The central atom is \(N\) (less electronegative than \(O\) and \(F\)).
• Connect \(N\) to \(O\) and \(N\) to \(F\) with single bonds (\(-\)). Each single bond uses \(2\) electrons. So far, \(2\times2 = 4\) electrons are used.

Step3: Complete the octets

• For \(O\): It needs \(6\) more electrons (since it has \(2\) from the bond). So we add \(3\) lone - pairs (\(\cdot\cdot\)) to \(O\).
• For \(F\): It needs \(6\) more electrons (since it has \(2\) from the bond). So we add \(3\) lone - pairs (\(\cdot\cdot\)) to \(F\).
• For \(N\): After forming two single bonds (\(4\) electrons used in bonds), we have \(18-(4 + 6+6)=2\) electrons left. These \(2\) electrons form a lone - pair (\(\cdot\cdot\)) on \(N\).
• Also, to satisfy the octet rule better, we can form a double bond (\(=\)) between \(N\) and \(O\).

Answer:

Top blue square: \(N\) (with a lone - pair \(\cdot\cdot\) on \(N\)), middle blue square (connecting \(N\) and \(O\)): \(=\), bottom blue square: \(F\) (with three lone - pairs \(\cdot\cdot\) on \(F\)) and \(O\) has three lone - pairs \(\cdot\cdot\)