Answer:
A) Q + XZ = X + QZ is a single displacement reaction.
B) Q + Z = QZ is a synthesis reaction
C) QT = Q + T is a decomposition reaction
D) QT + XZ = QZ + XT is double replacement reaction.
Explanation:
A) Q + XZ = X + QZ
This is a single displacement reaction because it is one in which one element is substituted for another one in a compound. In this case X is substituted for Q.
B) Q + Z = QZ
This is a synthesis reaction because Q and z combine to form a single product QZ.
C) QT = Q + T
This is a decomposition reaction because the compound QT breaks down to form 2 simpler substances Q and T.
D) QT + XZ = QZ + XT
Thus is a double replacement reaction because QT and XZ have exchanged cations to form new compounds QZ and XT
Answer:
Explanation:
All three lighter boron trihalides, BX3 (X = F, Cl, Br), form stable adducts with common Lewis bases. Their relative Lewis acidities can be evaluated in terms of the relative exothermicities of the adduct-forming reaction. Such measurements have revealed the following sequence for the Lewis acidity: BF3 < BCl3 < BBr3 (in other words, BBr3 is the strongest Lewis acid).
This trend is commonly attributed to the degree of π-bonding in the planar boron trihalide that would be lost upon pyramidalization (the conversion of the trigonal planar geometry to a tetrahedral one) of the BX3 molecule, which follows this trend: BF3 > BCl3 > BBr3 (that is, BBr3 is the most easily pyramidalized). The criteria for evaluating the relative strength of π-bonding are not clear, however. One suggestion is that the F atom is small compared to the larger Cl and Br atoms, and the lone pair electron in the 2pzorbital of F is readily and easily donated, and overlaps with the empty 2pz orbital of boron. As a result, the [latex]\pi[/latex] donation of F is greater than that of Cl or Br. In an alternative explanation, the low Lewis acidity for BF3 is attributed to the relative weakness of the bond in the adducts F3B-L.
Iodine electron configuration is:
1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10 5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.
So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5
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Explanation:
