Question

decide whether these proposed lewis structures are reasonable. proposed lewis structure is the proposed lewis structure reasonable? h = br yes. no, it has the wrong number of valence electrons. the correct number is: no, it has the right number of valence electrons but doesnt satisfy the octet rule. the symbols of the problem atoms are: :f = o = f: yes. no, it has the wrong number of valence electrons. the correct number is: no, it has the right number of valence electrons but doesnt satisfy the octet rule. the symbols of the problem atoms are: :o: || h - c - h yes. no, it has the wrong number of valence electrons. the correct number is: no, it has the right number of valence electrons but doesnt satisfy the octet rule. the symbols of the problem atoms are: if two or more atoms of the same element dont satisfy the octet rule, just enter the chemical symbol as many times as necessary. for example, if two oxygen atoms dont satisfy the octet rule, enter o,o.

Explanation:

Step1: Calculate valence - electrons for HBr

Hydrogen (H) has 1 valence electron and bromine (Br) has 7 valence electrons. So the total number of valence electrons for HBr is \(1 + 7=8\). In the proposed structure \(H\equiv Br\), the number of electrons shown is not correct for a single - bond between H and Br. A triple - bond is incorrect for HBr. The correct Lewis structure should have a single bond between H and Br with 3 lone - pairs on Br. The correct number of valence electrons is 8.

Step2: Calculate valence - electrons for \(F = O = F\)

Fluorine (F) has 7 valence electrons and oxygen (O) has 6 valence electrons. The total number of valence electrons for \(F_2O\) is \(2\times7 + 6=20\). In the proposed structure \(F = O = F\), the number of electrons shown is consistent with 20 valence electrons. But oxygen has 10 electrons around it (4 non - bonding electrons and 6 bonding electrons in two double - bonds), which violates the octet rule. The problem atoms are O.

Step3: Calculate valence - electrons for \(H_2CO\)

Hydrogen (H) has 1 valence electron, carbon (C) has 4 valence electrons and oxygen (O) has 6 valence electrons. The total number of valence electrons for \(H_2CO\) is \(2\times1+4 + 6=12\). The proposed structure shows the correct number of valence electrons and all atoms (H has 2 electrons, C has 8 electrons and O has 8 electrons) satisfy the octet rule (H follows the duet rule).

Answer:

For \(H\equiv Br\): No, it has the wrong number of valence electrons. The correct number is: 8
For \(F = O = F\): No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are: O
For \(H_2CO\): Yes.