(03.02 MC) A student made the Lewis dot diagram of a compound as shown. Ca is written with two dots shown on its top. Cl is writ
ten on the right side of Ca. Cl has seven dots around it. An arrow is shown from one dot on Ca toward the vacant space around Cl. Similarly, another arrow is shown from the other dot on Ca toward the vacant space around Cl. On the top, the art title is written as Students Lewis Dot Model. What is the error in the Lewis dot diagram? The Ca atom should transfer only one electron to Cl because the formula is CaCl. The number of dots around Ca should be one because Ca has to transfer only one electron to Cl. The Ca atom should transfer only one electron to Cl, and the second electron should be transferred to another Cl atom. The Cl atom should transfer two electrons to Ca because it has more electrons than Ca, so the formula becomes CaCl2.
2 answers:
The answer is The Ca atom should transfer only one electron to Cl, and the second electron should be transferred to another Cl atom.
the lewis dot structure of calcium chloride is as shown in the image attached:
one atom of Ca is giving its two electrons to chlorine atoms so that it makes ionic bond with chlorine, it gets +2 charge, and two chlorine atoms will get -1 negative charge respectively that is overall the compound is neutral.
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I have attached an image of the IR spectrum required to answer this question.
Looking at the IR, we can look for any clear major stretches that stand out. Immediately, looking at the spectrum, we see an intense stretch at around 1700 cm⁻¹. A stretch at this frequency is due to the C=O stretch of a carbonyl. Therefore, we know our answer must contain a carbonyl, so it could still be a ketone, aldehyde, carboxylic, ester, acid chloride or amide. However, if we look in the 3000 range of the spectrum, we see some unique pair of peaks at 2900 and 2700. These two peaks are characteristic of the sp² C-H stretch of the aldehyde.
Therefore, we can already conclude that this spectrum is due to an aldehyde based on the carbonyl stretch and the accompanying sp² C-H stretch.
Looking at the IR, we can look for any clear major stretches that stand out. Immediately, looking at the spectrum, we see an intense stretch at around 1700 cm⁻¹. A stretch at this frequency is due to the C=O stretch of a carbonyl. Therefore, we know our answer must contain a carbonyl, so it could still be a ketone, aldehyde, carboxylic, ester, acid chloride or amide. However, if we look in the 3000 range of the spectrum, we see some unique pair of peaks at 2900 and 2700. These two peaks are characteristic of the sp² C-H stretch of the aldehyde.
Therefore, we can already conclude that this spectrum is due to an aldehyde based on the carbonyl stretch and the accompanying sp² C-H stretch.